midterm

Required information Skip to question The temperature on a very warm summer day is 111°F. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is this temperature on the absolute (Kelvin) scale? The temperature is Numeric ResponseEdit Unavailable. not attempted, incorrect.K.

Explanation First convert the temperature in Fahrenheit to degrees Celsius and then to Kelvin. TC=(TF−32)×5/9 TC=((111−32)×5/9) °C=43.8889 °C TK =TC+273.15 TK =43.8889°C+273.15 TK =317.0389 K

Required information Skip to question 550 calories of heat are added to a gas. In addition, the gas expands and does 400 joules of work on its surroundings. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. How many joules does 550 calories represent? 550 calories represent Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation Heat in cal is converted to Joule using the following formula: QJ=Qcal×4.186=(550×4.186) J QJ= 2302.3 J

Required information Skip to question 550 calories of heat are added to a gas. In addition, the gas expands and does 400 joules of work on its surroundings. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the change in the internal energy of the gas in joules? The change in the internal energy of the gas is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation Heat in calorie is converted to Joule using the following formula: QJ=Qcal×4.186=(550×4.186) J QJ=2302.3 J The change in the internal energy of the gas is calculated using the following formula: ΔU=Q−W= 2302.3 J−400 J where Q is the amount of heat added or removed from a system, and W is the work by the gas. ΔU=1902.3 J

Required information Skip to question 550 calories of heat are added to a gas. In addition, the gas expands and does 400 joules of work on its surroundings. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the change in the internal energy of the gas in calories? The change in the internal energy is Numeric ResponseEdit Unavailable. not attempted, incorrect.cal.

Explanation Heat in calorie is converted to Joules using the following formula: QJ=Qcal×4.186=(550×4.186) J QJ=2302.3 J The change in the internal energy of the gas is calculated using the following formula: ΔU=Q−W= 2302.3 J−400 J where Q is the heat applied or removed to the gas, and W is the work done by the gas. ΔU=1902.3 J Now the change in the internal energy of the gas obtained in joules is converted to calories as follows: E(cal)=E(J)/4.186 Substituting the value in the equation, we get E(cal)=1902.3/4.186=454.4434 cal

Required information Skip to question At the low point in its swing, a pendulum bob with a mass of 0.15 kg has a velocity of 3 m/s. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is its kinetic energy at the high point? The kinetic energy at the high point is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation Since the velocity at the highest point is zero, the kinetic energy is zero.

If 80 cal of heat is added to a system, how much energy has been added in joules? The energy added to the system is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The amount of heat in calories is converted to energy in joules using the following formula: EJ=4.186×Hcal where EJ is the energy in joules and Hcal is the amount of heat in calories added to the system. EJ=4.186×80 J EJ=334.88 J

How much heat must be added to 135 g of ice at 0°C to melt the ice completely? The latent heat of fusion of ice is 80 cal/g. The amount of heat required is Numeric ResponseEdit Unavailable. not attempted, incorrect. cal.

Explanation The amount of heat required is calculated using the following formula: Q=mLf where m is the mass of the ice and Lf is the latent heat of fusion of ice. Q=135 g×80 cal/g Q=10800 cal

Required information Skip to question Sixty grams of water is at an initial temperature of 26 °C. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Calculate the additional heat required to completely convert the 100 ºC water to steam. (The latent heat of vaporization of water is 540 cal/g.) The additional heat required to completely convert the 100 ºC water to steam is Numeric ResponseEdit Unavailable. not attempted, incorrect.kcal.

Explanation The amount of heat that should be added to water is calculated using the following formula: Q=mLv where m is the mass of water, and Lv is the latent heat of vaporization of water. Q=60 g×540 cal/g=32400 cal

Required information Skip to question A net force of 100 N accelerates a 4-kg mass over a distance of 23 m. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the increase in kinetic energy of the mass? The increase in kinetic energy of the mass is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The amount of work done is also represented as the change in kinetic energy. The work done is calculated as follows: W=F×d W=100 N×23 m=2300 J Therefore, the increase in kinetic energy is equal to the work done, that is, 2300 J.

Required information Skip to question A 0.3-kg ball has a velocity of 42 m/s. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. How much work must be done on the ball to stop it? The work required to stop the ball is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The amount of work required to stop the ball is equal to the amount of work done to make the kinetic energy zero. The initial kinetic energy is calculated using the formula: Kinetic energy=1/2mvsquared For the given mass and velocity of the ball, the initial kinetic energy is calculated as follows: 1/2 mvsquared=1/2×0.3 kg×(42 m/s)squared=264.6 J Thus, the amount of work required is -264.6 J. The negative sign indicates that the required force should oppose the motion.

Required information Skip to question A uniform disk with a mass of 7 kg and a radius of 0.7 m is rotating with a rotational velocity of 15 rad/s. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the angular momentum of the disk? The angular momentum of the disk is Numeric ResponseEdit Unavailable. not attempted, incorrect.kg·m2/s.

Explanation The angular momentum of the disk is calculated using the following formula: L= Iω= 1.715 kg⋅msquared×(15 rad/s) = 25.725 kg⋅m2/sL= Iω= 1.715 kg⋅m2×(15 rad/s) = 25.725 kg⋅m2/s where I is the rotational inertia, and ω is the rotational velocity.

Required information Skip to question Suppose a disk rotates through 20 revolutions in 5 seconds. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is its average rotational velocity in rad/s? The average rotational velocity is Numeric ResponseEdit Unavailable. not attempted, incorrect.rad/s.

Explanation The average rotational velocity is calculated as follows: ωavg = θ/t=125.6637rad/5 s=25.1327 rad/s

How many times larger is the centripetal acceleration for a car rounding a curve at 90 MPH than for one rounding the same curve at 20 MPH? The centripetal acceleration is Numeric ResponseEdit Unavailable. not attempted, incorrect.times larger for the car rounding the curve at 90 MPH

Explanation The centripetal acceleration for the car rounding at 20 MPH is determined using the following formula: ac1=vsquared1/r ac1=(20 ×0.44704 m/s)squared/r=79.9379/r m/s2 where ac1 is the centripetal acceleration, v1 is the velocity, and r is the radius. The centripetal acceleration for the car rounding at 90 MPH is determined using the following formula: ac2=vsquared2/r ac2=(90 ×0.44704 m/s)squared/r=1618.7426/r m/s2 The ratio of the centripetal forces gives the factor by which the acceleration changes: ac2/ac1=1618.7426 m/s2/r /79.9397m/s2/r ac2/ac1=1618.7426/79.9397=20.2495 times

A ball is traveling at a constant speed of 6.5 m/s in a circle with a radius of 0.8 m. What is the centripetal acceleration of the ball? The centripetal acceleration of the ball is Numeric ResponseEdit Unavailable. not attempted, incorrect.m/s2.

Explanation The centripetal acceleration of the ball is determined using the following formula: ac=vsquaredxr ac=(6.5m/s)2/0.8 m=52.8125 m/s2 where ac is the centripetal acceleration, v is the velocity, and r is the radius.

Required information Skip to question A car with a mass of 1500 kg is moving around a curve with a radius of 45 m at a constant speed of 22 m/s. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Calculate the centripetal acceleration of the car. The centripetal acceleration of the car is Numeric ResponseEdit Unavailable. not attempted, incorrect.m/s2.

Explanation The centripetal acceleration of the car is determined using the following formula: ac=vsquared/r ac=(22 m/s)squared/45 m=10.7556 m/s2 where ac = centripetal acceleration, v = velocity, and r = radius.

A car rounds a curve with a radius of 40 m at a speed of 26 m/s. Calculate the centripetal acceleration of the car. The centripetal acceleration of the car is Numeric ResponseEdit Unavailable. not attempted, incorrect.m/s2.

Explanation The centripetal acceleration of the car is determined using the following formula: ac=vsquared/r ac=(26m/s)squared/40 m=16.9 m/s2 where ac is the centripetal acceleration, v is velocity, and r is the radius.

Required information Skip to question A Ferris wheel at a carnival has a radius of 8 m and turns so that the speed of the riders is 6.5 m/s. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Calculate the centripetal acceleration of the riders. The centripetal acceleration of the riders is Numeric ResponseEdit Unavailable. not attempted, incorrect.m/s2.

Explanation The centripetal acceleration of the riders is determined using the following formula: ac=vsquared/r ac=(6.5m/s)squared/8 m=5.2812 m/s2 where ac = centripetal acceleration, v = velocity, and r = radius.

Required information Skip to question Suppose a disk rotates through 20 revolutions in 5 seconds. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is its displacement in radians in this time? The disk's displacement is Numeric ResponseEdit Unavailable. not attempted, incorrect.radians.

Explanation The disk's displacement is calculated as follows: θ= 20 rev×2π rad/1 rev = 125.6637 rad

The volume of an ideal gas is increased from 2 m3 to 2.5 m3 while maintaining a constant pressure of 3600 Pa (1 Pa = 1 N/m2). If the initial temperature of the ideal gas is 480 K, calculate the final temperature. The final temperature is Numeric ResponseEdit Unavailable. not attempted, incorrect.K.

Explanation The final temperature of the ideal gas is calculated using the following formula: T2=V2/V1x T1 where V2 is the final volume, T1 is the initial temperature, and V1 is the initial volume. Substituting the value in the equation, we get T2=2.5 m3/2 m3x480 K= 600 K

If 900 cal of heat is added to 115 g of water initially at a temperature of 18 °C, what is the final temperature of the water? The final temperature of the water is Numeric ResponseEdit Unavailable. not attempted, incorrect.°C.

Explanation The final temperature of the water is calculated using the following formula: T2= Q/mc+T1 where Q is the heat added to the water, m is the mass of the water, c is the specific heat of water, and T1 is the initial temperature. T2=900 cal/115 g×1 cal/g°C+18°C T2=25.8261 °C

Required information Skip to question 1650 J of heat is added to 90 g of water at an initial temperature of 18°C. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the final temperature of the water? The final temperature of the water is Numeric ResponseEdit Unavailable. not attempted, incorrect.°C.

Explanation The final temperature of the water is calculated using the following formula: T2=Q/m+T1 where Q is the heat added to the water, m is the mass of the water, and T1 is the initial temperature. T2=(1650/4.186) cal/90 g+18 °C T2=22.3797 °C

Two masses are attracted by a gravitational force of 8.6 N. What will the force of attraction be if the distance between the two masses is quadrupled? The force of attraction will be Numeric ResponseEdit Unavailable. not attempted, incorrect.N.

Explanation The force of attraction is determined using Newton's law of universal gravitation: F(force at four times the distance)=Gm1m2/(4d)squared=Gm1m2/16dsquared F(force at four times the distance)=8.6 N/16=0.5375 N where m1 and m2 are the masses of two forces, d is the distance, and G is the universal gravitational constant.

Two masses are attracted by a gravitational force of 0.25 N. What will the force of attraction be if the distance between the two masses is halved? The force of attraction will be Numeric ResponseEdit Unavailable. not attempted, incorrect.N.

Explanation The force of attraction is determined using Newton's law of universal gravitation: F(force at half the distance)=Gm1m2(1/2d)squared=Gm1m2/0.25 dsquared F (force at half the distance)=0.25 N/0.25=1 N where m1 and m2 are the masses of two forces, d is the distance, and G is the universal gravitational constant.

Two 700-kg masses (1543 lb) are separated by a distance of 68 m. Using Newton's law of gravitation, find the magnitude of the gravitational force exerted by one mass on the other. (Use G = 6.67 × 10-11 N·m2/kg2.) The magnitude of the gravitational force exerted by one mass on the other is Numeric ResponseEdit Unavailable. not attempted, incorrect.× 10-9 N.

Explanation The force of attraction is determined using Newton's law of universal gravitation: F=Gm1m2/dsquared F=(700 kg × 700 kg) ×(6.67× 10−11 Nm2/kg2)/(68 m)2=7.0681×10−9 N where m1 and m2 are the masses of two objects, d is the distance between them, and G is the universal gravitational constant.

The acceleration of gravity on the surface of Jupiter is 25 m/s2. What is the weight on Jupiter of a woman whose weight on Earth is 130 lb? The weight of the woman on Jupiter is Numeric ResponseEdit Unavailable. not attempted, incorrect.lb.

Explanation The gravitational factor on Jupiter is determined as follows: acceleration of gravity on the surface of Jupiter/acceleration of gravity on the surface of Earth=25 m/s2/9.8 m/s2=2.55 Thus, the weight of a person on Jupiter is 2.55 times the value of his weight on Earth. Therefore, the weight of the woman on Jupiter is calculated as follows: W=130 lb × 2.55=331.5 lb

The acceleration due to gravity at the surface of the moon is 1.63 m/s2. What is the weight of an astronaut standing on the moon whose weight on Earth is 174 lb? The weight of the astronaut on the moon is Numeric ResponseEdit Unavailable. not attempted, incorrect.lb.

Explanation The gravitational factor on the moon is determined as follows: acceleration of gravity on the surface of the moon/acceleration of gravity on the surface of the Earth=1.63 m/s2/9.8 m/s2=0.166 The weight of a person on the moon is one-sixth (1/6 = 0.166) of the value of his weight on the Earth. Therefore, the weight of the astronaut on the moon is calculated as follows: 174 lb/6=29 lb

Dylan has a weight of 600 N when he is standing on the surface of the Earth. What would his weight (the gravitational force due to the Earth) be if he tripled his distance from the center of the Earth by flying in a spacecraft? His weight would be Numeric ResponseEdit Unavailable. not attempted, incorrect.N when the distance from the center of the Earth is tripled.

Explanation The gravitational force (weight of the person) is determined using Newton's law of universal gravitation: F(force at thrice the distance)=Gm1m2/(3d)squared=Gm1m2/9dsquared F(force at thrice the distance)=600 N9=66.6667 N where m1 and m2 are the masses of the two forces, d is the distance, and G is the universal gravitational constant.

Required information Skip to question A box with a mass of 7 kg is lifted (without acceleration) through a height of 1.8 m, in order to place it upon the shelf of a closet. The value of acceleration due to gravity g = 9.8 m/s2. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. How much work must a person do to lift the box to this position? The work done to lift the box is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The gravitational potential energy is equal to the work done to lift the box. For the given mass of the box and the height, the potential energy is calculated as follows: mgh=7 kg×9.8 m/s2×1.8 m=123.48 J Thus, the work done to lift the box to this position is equal to the potential energy, that is, 123.48 J.

How much heat is required to raise the temperature of 100 g of water from 12°C to 88°C? The specific heat capacity of water is 1 cal/g·°C. The heat required is Numeric ResponseEdit Unavailable. not attempted, incorrect.cal.

Explanation The heat required to raise the temperature of 100 g of water from 12°C to 88°C is calculated using the following formula: Q=mcΔT where m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature. Q=(100 g)×(1 cal/g⋅°C)×(88 °C−12 °C) Q = 7600 cal

Calculate the amount of heat that must be removed from an 700-g block of aluminum to lower its temperature from 160°C to 40°C. The specific heat capacity of aluminum is 0.215 cal/g·°C. The amount of heat that must be removed from the block of aluminum is Numeric ResponseEdit Unavailable. not attempted, incorrect.cal.

Explanation The heat that must be removed from the block of aluminum is calculated using the following formula: Q=mcΔT where m is the mass of the aluminum, c is the specific heat of aluminum, and ΔT is the change in temperature. Q= (700 g)×(0.215 cal/g ⋅°C)×(160°C−40 °C) Q=18060 cal

Required information Skip to question A 15-kg rock is going to be moved either vertically or horizontally. Consider the acceleration due to gravity as 9.8 m/s2. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the increase in the kinetic energy of the rock when accelerating horizontally it to a speed of 6 m/s? The increase in kinetic energy is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The increase in kinetic energy is given as KE=1/2mvsquared =1/2×15 kg×(6 m/s)squared=270 J

Required information Skip to question A sled and rider with a combined mass of 160 kg are at the top of a hill of height 13.5 m above the level ground below. The sled is given a push providing an initial kinetic energy of 1650 J at the top of the hill. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What will be the kinetic energy of the sled and rider at the bottom of the hill, if friction can be ignored? The kinetic energy of the sled and rider at the bottom of the hill is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The kinetic energy at the bottom of the hill is equal to the total mechanical energy at the top of the hill since the friction is not considered. The total mechanical energy is the sum of the potential energy and the kinetic energy of a system. The potential energy is calculated as follows: Potential energy=mgh=160 kg×9.8 m/s2×13.5 m Potential energy=21168 J The kinetic energy of the system is1650 J Therefore,the mechanical energy is calculated as follows: 21168 J + 1650 J = 22818 J The kinetic energy at the bottom of the hill is 22818 j

Required information Skip to question At the low point in its swing, a pendulum bob with a mass of 0.15 kg has a velocity of 3 m/s. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is its kinetic energy at the low point? The kinetic energy at the low point is J.

Explanation The kinetic energy at the low point is calculated as follows: Kinetic energy=1/2mvsquared=1/2×0.15 kg×(3 m/s)squared=0.675 J

A roller-coaster car has a potential energy of 400 kJ and a kinetic energy of 130 kJ at point A in its travel. At the low point of the ride, the potential energy is zero, and 60 kJ of heat has been generated by friction since it left point A. What is the kinetic energy of the roller coaster at this low point? The kinetic energy of the roller coaster at its low point is Numeric ResponseEdit Unavailable. not attempted, incorrect.kJ.

Explanation The kinetic energy at the lowest point is equal to the initial mechanical energy minus the energy lost to friction. The initial mechanical energy is the sum of the initial potential energy and the initial kinetic energy. Some of this energy is lost to friction as the car moves. Therefore, the kinetic energy at the lowest point is calculated as follows: 400 kJ + 130 kJ - 60 kJ = 470 kJ

Required information Skip to question A 0.3-kg ball has a velocity of 42 m/s. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the kinetic energy of the ball? The kinetic energy of the ball is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The kinetic energy is calculated using the following formula: Kinetic energy=1/2mvsquared For the given mass and velocity of the ball, the kinetic energy is calculated as follows: 1/2 mvsquared=1/2×0.3 kg×(42 m/s)squared=264.6 J

A 0.35-kg ball moving in a circle at the end of a string has a centripetal acceleration of 10 m/s2. Determine the magnitude of the centripetal force exerted by the string on the ball to produce this acceleration. The magnitude of the centripetal force is Numeric ResponseEdit Unavailable. not attempted, incorrect.N

Explanation The magnitude of the centripetal force is determined using the following formula: Fnet=ma Fnet= (0.35 kg)(10 m/s2) = 3.5 N where a = centripetal acceleration, m = mass of the ball, and Fnet = magnitude of the centripetal force.

Required information Skip to question A car with a mass of 1500 kg is moving around a curve with a radius of 45 m at a constant speed of 22 m/s. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Determine the magnitude of the force required to produce this centripetal acceleration. The magnitude of the force is Numeric ResponseEdit Unavailable. not attempted, incorrect.N.

Explanation The magnitude of the force is determined using the following formula: Fnet=ma Fnet=(1500kg)(10.7556 m/s2) = 16133.3333 N where a = centripetal acceleration, m = mass of the car, and Fnet = magnitude of force.

Required information Skip to question A Ferris wheel at a carnival has a radius of 8 m and turns so that the speed of the riders is 6.5 m/s. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the magnitude of the net force required to produce this centripetal acceleration for a rider with a mass of 75 kg? The magnitude of the force is Numeric ResponseEdit Unavailable. not attempted, incorrect.N.

Explanation The magnitude of the net force is determined using the following formula: Fnet= ma Fnet= (75 kg)(5.28125 m/s2) = 396.0938 N where a = centripetal acceleration, m = mass of the ball, and Fnet = magnitude of the net force.

Required information Skip to question A torque of 80 N·m producing a counterclockwise rotation is applied to a wheel about its axle. A frictional torque of 14 N·m acts at the axle. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the net torque about the axle of the wheel? The net torque about the axle of the wheel is Numeric ResponseEdit Unavailable. not attempted, incorrect.N·m.

Explanation The net torque about the axle of the wheel is calculated using the following formula: τnet= τapplied+τfrictional= 80 N⋅m+(−14 N⋅m) = 66 N⋅m

Required information Skip to question Two forces are applied to a merry-go-round with a radius of 1.6 m as shown in the following diagram. One force has a magnitude of 90 N and the other a magnitude of 40 N. Take the torque caused by the 90-N force to be positive. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Calculate the net torque acting on the merry-go-round. The net torque acting on the merry-go-round is Numeric ResponseEdit Unavailable. not attempted, incorrect.N·m.

Explanation The net torque acting on the merry-go-round is calculated using the following formula: τtotal = τ90 N+τ40 N = (144 N⋅m)+(-64 N⋅m) = 80 N⋅m

A merry-go-round starts from rest and accelerates at a constant rate of 1.3 rev/s2. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. How many revolutions does the merry-go-round make in 6 s? The number of revolutions made by the merry-go-round is Numeric ResponseEdit Unavailable. not attempted, incorrect..

Explanation The number of revolutions made by the merry-go-round is calculated using the following formula: θ = ω0t+1/2αt2 = 0+1/2(1.3 rev/s2)(6 s)squared = 23.4 rev

Required information Skip to question A bicycle wheel is rotationally accelerated at a constant rate of 2.2 rev/s2. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Through how many revolutions does it turn in 5 s? The number of revolutions made by the wheel is Numeric ResponseEdit Unavailable. not attempted, incorrect.

Explanation The number of revolutions made by the wheel is calculated using the following formula: θ= ω0t+1/2αtsquared= 0+1/2(2.2 rev/s2×(5 s)squared)= 27.5 rev

Required information Skip to question When one of the authors was a teenager, the rate of rotation for popular music records on a record player was 45 RPM. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Determine the number of revolutions the record makes in 5 s. The number of revolutions made in 5 s is Numeric ResponseEdit Unavailable. not attempted, incorrect..

Explanation The number of revolutions the record makes in 5 s is calculated using the following formula: Number of revolutions = (45 RPM)×1 min/60 s×5 s = 3.75

Required information Skip to question At the low point in its swing, a pendulum bob with a mass of 0.15 kg has a velocity of 3 m/s. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is its potential energy at the high point (assuming the potential energy at the low point was zero)? The potential energy at the high point is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The potential energy at the high point is equal to the kinetic energy at the lowest point. Therefore, Potential energy=1/2mvsquared=1/2×0.15 kg×(3 m/s)squared=0.675 J

Required information Skip to question A sled and rider with a combined mass of 160 kg are at the top of a hill of height 13.5 m above the level ground below. The sled is given a push providing an initial kinetic energy of 1650 J at the top of the hill. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Choosing a reference level at the bottom of the hill, what is the potential energy of the sled and rider at the top of the hill? The potential energy of the sled and rider at the top of the hill is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The potential energy is calculated as follows: Potential energy=mgh=160 kg×9.8 m/s2×13.5 m Potential energy=21168 J

Required information Skip to question A box with a mass of 7 kg is lifted (without acceleration) through a height of 1.8 m, in order to place it upon the shelf of a closet. The value of acceleration due to gravity g = 9.8 m/s2. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the increase in potential energy of the box? The increase in potential energy of the box is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The potential energy is calculated using the following formula: Potential energy = mgh For the given mass of the box and the height, the potential energy is calculated as follows: mgh=7 kg×9.8 m/s2×1.8 m=123.48 J

To stretch a spring a distance of 0.3 m from the equilibrium position, 145 J of work is done. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the value of the spring constant k? The value of the spring constant k is Numeric ResponseEdit Unavailable. not attempted, incorrect.N/m.

Explanation The potential energy is calculated using the following formula: Potential energy=1/2kx2Potential energy=1/2kxsquared where k is the spring constant, and x is the distance traversed by the spring. Since the potential energy is equal to the work done at the equilibrium position, the spring constant is calculated as follows: k=2×145 J/(0.3 m)squared=3222.2222 N/m

A coil spring in an off-road truck with a spring constant k of 87.6 kN/m (87,600 N/m) is compressed a distance of 12.2 cm (0.122 m) from its original unstretched position. What is the increase in potential energy of the spring? The increase in the potential energy of the spring is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The potential energy of the spring is calculated using the following formula: Potential energy=1/2kxsquared where k is the spring constant, and x is the distance the spring is stretched. The increase in potential energy is calculated as follows: 1/2kxsquared=1/2×87600 N/m×(0.122 m)squared=651.9192 J

A ball traveling in a circle with a constant speed of 15 m/s has a centripetal acceleration of 20 m/s2. What is the radius of the circle? The radius of the circle is Numeric ResponseEdit Unavailable. not attempted, incorrect.m.

Explanation The radius of the circle is determined using the following formula: ac=vsquared/r r= vsquared/ac r=(15m/s)squared/20 m/s2=11.25 mr=(15m/s)220 m/s2=11.25 m where ac is the centripetal acceleration, v is the velocity, and r is the radius.

Required information Skip to question 1650 J of heat is added to 90 g of water at an initial temperature of 18°C. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. How much energy is this in calories? The equivalent energy is Numeric ResponseEdit Unavailable. not attempted, incorrect.cal.

Explanation The relationship between joules and calories is as follows: E(cal)=E(J) /4.186 The equivalent energy is calculated using the formula as follows: E(cal)=E(J)/4.186=(1650/4.186) cal E(cal)=394.171 cal

Required information Skip to question Sixty grams of water is at an initial temperature of 26 °C. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Calculate the heat required to raise the temperature to its boiling point. The required amount of heat is Numeric ResponseEdit Unavailable. not attempted, incorrect.cal.

Explanation The required amount of heat is calculated using the following formula: Q=mcΔT where m is the mass of water, c is the specific heat of water, and ΔT is the change in temperature. Q=60 g×(1 cal/g⋅°C)×(100°C−26°C) Q = 4440 cal

A weight of 30 N is located at a distance of 8 cm from the fulcrum of a simple balance beam. At what distance from the fulcrum should a weight of 25 N be placed on the opposite side in order to balance the system? The required distance from the fulcrum to balance the system is Numeric ResponseEdit Unavailable. not attempted, incorrect.cm.

Explanation The required distance from the fulcrum to balance the system is calculated as follows: d2 = F1⋅d1/F2 = 30 N⋅ 8 cm/25 N = 9.6 cm

The rotational velocity of a spinning disk decreases from 46 rev/s to 3 rev/s in a time of 15 s. What is the rotational acceleration of the disk? Take the initial and final directions of rotation to be positive. The rotational acceleration of the disk is Numeric ResponseEdit Unavailable. not attempted, incorrect.rev/s2.

Explanation The rotational acceleration of the disk is calculated using the following formula: α= ω− ω0/t = 3 rev/s − 46 rev/s/15 s = -2.8667 rev/s2

The rotational velocity of a merry-go-round increases at a constant rate from 3.1 rad/s to 23.1 rad/s in a time of 14 s. What is the rotational acceleration of the merry-go-round? The rotational acceleration of the merry-go-round is Numeric ResponseEdit Unavailable. not attempted, incorrect.rad/s2.

Explanation The rotational acceleration of the merry-go-round is calculated using the following formula: α=Δω/t=23.1 rad/s−3.1 rad/s/14 s α=1.4286 rad/s2

Required information Skip to question A uniform disk with a mass of 7 kg and a radius of 0.7 m is rotating with a rotational velocity of 15 rad/s. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the rotational inertia of the disk? The rotational inertia of the uniform disk is Numeric ResponseEdit Unavailable. not attempted, incorrect.kg·m2.

Explanation The rotational inertia of the uniform disk is calculated using the following formula: I=1/2mrsquared=1/2 (7 kg×(0.7 m)squared) = 1.715 kg⋅m2 I=12mr2=12 (7 kg×(0.7 m)2) = 1.715 kg⋅m2 where m is the mass of the disk, and r is the radius of the disk.

Required information Skip to question A torque of 80 N·m producing a counterclockwise rotation is applied to a wheel about its axle. A frictional torque of 14 N·m acts at the axle. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. If the wheel is observed to accelerate at the rate of 6 rad/s2 under the influence of these torques, what is the rotational inertia of the wheel? The rotational inertia of the wheel is Numeric ResponseEdit Unavailable. not attempted, incorrect.kg·m2.

Explanation The rotational inertia of the wheel is calculated using the following formula: I= τnetα = 66 N⋅m/6 rad/s2 = 11 kg⋅m2

Required information Skip to question When one of the authors was a teenager, the rate of rotation for popular music records on a record player was 45 RPM. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Express the rotational velocity in rev/s. The rotational velocity in rev/s is Numeric ResponseEdit Unavailable. not attempted, incorrect..

Explanation The rotational velocity in RPM is converted to rev/s as follows: ω=(45 RPM)×1min/60 s = 0.75 rev/s

Required information Skip to question A merry-go-round is rotating at the rate of 52 rev/min. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Express the rotational velocity in rad/s. The rotational velocity in rad/s is Numeric ResponseEdit Unavailable. not attempted, incorrect..

Explanation The rotational velocity in rev/min is converted to rad/s as follows: ω= 52 rev/min×1min ×2⋅π rad/60 s ×1 rev = 5.4454 rad/s

Required information Skip to question A merry-go-round is rotating at the rate of 52 rev/min. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Express the rotational velocity in rev/s. The rotational velocity in rev/s is Numeric ResponseEdit Unavailable. not attempted, incorrect..

Explanation The rotational velocity in rev/min is converted to rev/s as follows: ω= 52 rev/min×1 min /60 s = 0.8667 rev/s

A merry-go-round starts from rest and accelerates at a constant rate of 1.3 rev/s2. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is its rotational velocity after 6 s? The rotational velocity of the merry-go-round is Numeric ResponseEdit Unavailable. not attempted, incorrect.rev/s.

Explanation The rotational velocity of the merry-go-round is calculated using the following formula: ω = ω0+αt = 0+(1.3 rev/s2)(6 s) = 7.8 rev/s

Required information Skip to question The temperature on a very warm summer day is 111°F. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the temperature in degrees Celsius? The temperature is Numeric ResponseEdit Unavailable. not attempted, incorrect.ºC.

Explanation The temperature given in degrees Fahrenheit is converted to degrees Celsius as follows: TC=(TF−32)×5/9 where TF is the temperature given in degrees Fahrenheit. ​ Substituting the value in the equation, we get TC=((111−32)×5/9) °C=43.8889 °C

Required information Skip to question Two forces are applied to a merry-go-round with a radius of 1.6 m as shown in the following diagram. One force has a magnitude of 90 N and the other a magnitude of 40 N. Take the torque caused by the 90-N force to be positive. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the torque about the axle of the merry-go-round due to the 40-N force? The torque about the axle of the merry-go-round due to the 40-N force is Numeric ResponseEdit Unavailable. not attempted, incorrect.N·m.

Explanation The torque about the axle of the merry-go-round due to the 40-N force is calculated using the following formula: τ = F⋅R =−40 N×1.6 m =-64 N⋅m (Negative sign indicates the direction of the resulting torque)

Required information Skip to question Two forces are applied to a merry-go-round with a radius of 1.6 m as shown in the following diagram. One force has a magnitude of 90 N and the other a magnitude of 40 N. Take the torque caused by the 90-N force to be positive. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the torque about the axle of the merry-go-round due to the 90-N force? The torque about the axle of the merry-go-round due to the 90-N force is Numeric ResponseEdit Unavailable. not attempted, incorrect.N·m.

Explanation The torque about the axle of the merry-go-round due to the 90-N force is calculated using the following formula: τ = F⋅R =90 N⋅1.6 m = 144 N⋅m

A force of 80 N is applied at the end of a wrench handle that is 24 cm long. The force is applied in a direction perpendicular to the handle. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Calculate the torque if the force is applied halfway up the handle instead of at the end. The torque on the nut is Numeric ResponseEdit Unavailable. not attempted, incorrect.N·m.

Explanation The torque applied to the nut by the wrench is calculated using the following formula: τ = F⋅l = 80 N⋅(1/2×24 cm×1 m/100 cm)=9.6 N⋅m

A force of 80 N is applied at the end of a wrench handle that is 24 cm long. The force is applied in a direction perpendicular to the handle. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the torque applied to the nut by the wrench? The torque applied to the nut by the wrench is Numeric ResponseEdit Unavailable. not attempted, incorrect.N·m.

Explanation The torque applied to the nut by the wrench is calculated using the following formula: τ = F⋅l = 80 N⋅(24 cm×1 m/100 cm)= 19.2 N⋅m

Required information Skip to question A sled and rider with a combined mass of 160 kg are at the top of a hill of height 13.5 m above the level ground below. The sled is given a push providing an initial kinetic energy of 1650 J at the top of the hill. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. After the push, what is the total mechanical energy of the sled and rider at the top of the hill? The total mechanical energy of the sled and rider at the top of the hill is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The total mechanical energy is the sum of the potential energy and the kinetic energy of a system. The potential energy is calculated using the given formula as follows: Potential energy=mgh=160 kg×9.8 m/s2×13.5 m Potential energy=21168 J The kinetic energy of the system is1650 J Therefore, the mechanical energy is calculated as follows: 21168 J + 1650 J = 22818 J

Required information Skip to question A force of 130 N is used to drag a crate a distance of d = 5 m across a floor. The force is directed at an angle upward from the crate so that the vertical component of the force is 120 N and the horizontal component is 50 N as shown in the diagram. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the total work done by the 130-N force? The total work done by the 130-N force is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The total work done is calculated as the sum of the work done by both the horizontal and the vertical components of the applied force. The work done by the horizontal component of force is calculated as follows: Whorizontal=F×d=50 N×5 m=250 J Since the vertical component of the force is zero, the calculation of the total work done is as follows: Wtotal=250 J+0 J=250 J

Required information Skip to question A rope applies a horizontal force of 230 N to pull a crate a distance of 7 m across the floor. A frictional force of 120 N opposes this motion. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the total work done on the crate? The total work done on the crate is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The total work done on the crate is the sum of the work done by the applied force on the rope and the work done by the frictional force. The work done by the applied force on the rope is calculated using the following formula: W=F×d=230 N × 7 m W=1610 J The frictional force works in a direction opposite to that of the applied force. The work done by the frictional force is calculated as follows: Wfriction=Ffriction×d=−120 N×7 m Wfriction=−840 J The total work done is calculated as follows: Wtotal=1610 J+(- 840 J)=770 J

A weight of 6 N is located 12 cm from the fulcrum on the beam of a simple balance. Determine the weight that should be placed at a point 5 cm from the fulcrum on the opposite side in order to balance the system. The weight that should be placed at a point 5 cm from the fulcrum on the opposite side is Numeric ResponseEdit Unavailable. not attempted, incorrect.N.

Explanation The weight that should be placed at a point 5 cm from the fulcrum on the opposite side is calculated as follows: F2 = F1⋅d1/d2 = (6 N)⋅12 cm/5 cm = 14.4 N

Required information Skip to question A bicycle wheel is rotationally accelerated at a constant rate of 2.2 rev/s2. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. If it starts from rest, what is its rotational velocity after 5 s? The wheel's rotational velocity is Numeric ResponseEdit Unavailable. not attempted, incorrect.rev/s.

Explanation The wheel's rotational velocity is calculated using the following formula: ω= ω0+αt= 0+(2.2 rev/s2×5 s) ω= 11 rev/s

Required information Skip to question A rope applies a horizontal force of 230 N to pull a crate a distance of 7 m across the floor. A frictional force of 120 N opposes this motion. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the work done by the frictional force? The magnitude of the work done by the frictional force is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The work done by a given force is the product of the component of the force along the line of motion of the object multiplied by the distance moved by the object under the influence of the force. The frictional force works in a direction opposite to that of the applied force. The work done by the frictional force is calculated as follows: Wfriction=Ffriction×d=−120 N×7 m |Wfriction|=840 J

Required information Skip to question A force of 130 N is used to drag a crate a distance of d = 5 m across a floor. The force is directed at an angle upward from the crate so that the vertical component of the force is 120 N and the horizontal component is 50 N as shown in the diagram. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the work done by the horizontal component of the force? The work done by the horizontal component of the force is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The work done by a given force is the product of the component of the force along the line of motion of the object multiplied by the distance moved by the object under the influence of the force. The work done by the horizontal component of force is calculated as follows: Whorizontal=F×d=50 N×5 m=250 J

A horizontally directed force of 70-N is used to pull a box through a distance of 2.1 m across a tabletop. How much work is done by the 70-N force? The work done by the 70-N force is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The work done by a given force is the product of the component of the force along the line of motion of the object multiplied by the distance moved by the object under the influence of the force. The work done in joules is calculated as follows: W=F×d=70 N×2.1 m W=147 J

Required information Skip to question A net force of 100 N accelerates a 4-kg mass over a distance of 23 m. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the work done by this net force? The work done by the net force is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The work done by a given force is the product of the component of the force along the line of motion of the object multiplied by the distance moved by the object under the influence of the force. The work done is calculated as follows: W=F×d W=100 N×23 m=2300 J

Required information Skip to question A rope applies a horizontal force of 230 N to pull a crate a distance of 7 m across the floor. A frictional force of 120 N opposes this motion. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the work done by the force applied by the rope? The work done is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The work done by a given force is the product of the component of the force along the line of motion of the object multiplied by the distance moved by the object under the influence of the force. The work done is calculated as follows: W=F×d=230 N × 7 m W=1610 J

A woman does 250 J of work to move a table 1.8 m across the floor. What is the magnitude of the force applied by the woman to the table if this force is applied in the horizontal direction? The magnitude of the force applied by the woman is Numeric ResponseEdit Unavailable. not attempted, incorrect.N.

Explanation The work done by a given force is the product of the component of the force along the line of motion of the object multiplied by the distance moved by the object under the influence of the force. The work done is calculated using the following formula: W=F×dW=F×d The force is calculated as follows: F=W/d=250 J/1.8 mF=W/d=250 J/1.8 m F=138.8889 N

Required information Skip to question A force of 130 N is used to drag a crate a distance of d = 5 m across a floor. The force is directed at an angle upward from the crate so that the vertical component of the force is 120 N and the horizontal component is 50 N as shown in the diagram. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the work done by the vertical component of the force? The work done by the vertical component of the force is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The work done by a given force is the product of the component of the force along the line of motion of the object multiplied by the distance moved by the object under the influence of the force. The distance moved vertically is zero. Wvertical=F×d=50 N×0 m=0 J

The volume of an ideal gas is increased from 1.8 m3 to 2.5 m3 while maintaining a constant pressure of 3600 Pa (1 Pa = 1 N/m2). Calculate the amount of work done by the gas during the expansion. The amount of work done by the gas during the expansion is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation The work done by the gas is calculated using the following formula: W=PΔV=3600 Pa×(2.5 m3−1.8 m3) W=2520 J

Required information Skip to question A 15-kg rock is going to be moved either vertically or horizontally. Consider the acceleration due to gravity as 9.8 m/s2. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the increase in the potential energy of the rock when lifting it to a height of 1.8 m without acceleration? The increase in potential energy of the rock is Numeric ResponseEdit Unavailable. not attempted, incorrect.J.

Explanation We can obtain the potential energy by the following formula: PE=mgh=15 kg×1.8 m×9.8 m/s2=264.6 J

Required information Skip to question At the low point in its swing, a pendulum bob with a mass of 0.15 kg has a velocity of 3 m/s. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. How high will the bob swing above the low point before reversing direction? Ignore the air resistance. The bob swings a height of Numeric ResponseEdit Unavailable. not attempted, incorrect.m.

Explanation When the bob swings to its highest point, the potential energy is equal to the kinetic energy at the bottom of the swing. The kinetic energy at the low point is calculated as follows: Kinetic energy=12mv2=12×0.15 kg×(3 m/s)2=0.675 JKinetic energy=12mv2=12×0.15 kg×(3 m/s)2=0.675 J The potential energy at the high point is equal to the kinetic energy at the low point. Since the potential energy at the high point is given by mgh, it folllows that h=Kinetic energy/mg=0.675 J/0.15 kg×9.8 m/s2 h=0.4592 m

A horizontal force of 130 N used to push a chair across a room does 420 J of work. How far does the chair move in this process? The chair moves a distance of Numeric ResponseEdit Unavailable. not attempted, incorrect.m.

Explanation The work done by a given force is the product of the component of the force along the line of motion of the object multiplied by the distance that the object moves under the influence of the force. The work done is calculated using the formula: W=F×dW=F×d The distance through which the chair is moved is calculated as follows: d=W/F=420 J/130 N=3.2308 m

To stretch a spring a distance of 0.3 m from the equilibrium position, 145 J of work is done. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the increase in potential energy of the spring? The increase in potential energy of the spring is Numeric ResponseEdit Unavailable. not attempted, incorrect.J. Explanation

Once the spring has been stretched, the work done is equal to the change in potential energy. The potential energy is equal to 145 J.