Physics Questions 1-3

A Doppler study of a patient's aorta was used to determine the speed of blood flow at 0.40 m/s in the thoracic aorta and then at a point 0.4 m away, 0.20 m/s in the abdominal aorta. Assuming that the blood flow slowed at a constant rate, what is the rate that the blood slowed between the 2 points measured along the aorta? A) 0.15 m/s2 B) 0.30 m/s2 C) 1.50 m/s2 D) 3.00 m/s2 Ch 1

A We can tell from the units, that the question is referring to acceleration (or, to be more precise, since it is the rate of slowing, deceleration). 1) acceleration = (change in velocity)/(change in time) We know the change in velocity (Final - initial = 0.20 - 0.40 = -0.20 m/s), so now we need to determine the change in time. 2) If we can assume that the blood slowed in the aorta at a constant rate, then the average speed of blood flow was (0.40 + 0.20)/2 = 0.30 m/s over a distance of 0.4 m, thus (0.4 m)/(0.30 m/s) = 1.33 s. {Note: this is determined by either paying attention to the units (= dimensional analysis) or by knowing that velocity = distance/time therefore time = distance/velocity.} 3) acceleration = (change in velocity)/(change in time) = - (1/5)/(4/3) = -3/20 = -15/100 = -0.15. Thus there is a deceleration (a rate of slowing) of 0.15 m/s2. Side note: Whenever possible, avoid decimals and use fractions to the very last moment since they are easier to manipulate (i.e. you'll be able to work faster).

10) Again, consider Figure 1. The runner starts at position X = 0. What is the final position (as measured from X = 0) after she runs through the motion described by the graph, from t = 0 seconds to t = 5 seconds? A. 22.5 m B. 25 m C. 27.5 m D. 32.5 m

A triangle's area is given by: ½ base x height. (A) 1/2 (1.5)(10) = 7.5 m (C) ½ (1)(10) = 5 m (D) ½ (1)(-10) = -5 m A rectangle's area is given by: base x height (B) 10 x 1.5 = 15 m (A) + (B) + (C) + (D) = 7.5 + 15 + 5 - 5 = 27.5 - 5 = 22.5 m

3) A net force of 10 N accelerates an object at 5.0 m/s2. What net force would be required to accelerate the same object at 1.0 m/s2? A. 2.0 N B. 3.0 N C. 4.0 N D. 5.0 N

Your choice: A Correct choice: A Explanation: Using Newton's 2nd Law, F = ma, we get: m = F/a = 10/5 = 2 kg to calculate the mass of the object. Now we use F = ma = 2 kg x 1 m/s2 = 2 N to find the force.

Three little girls each pull on the same object with a 10 N force as shown below. The resultant force will be: 10 N up (F2) and 10 N down (F3) A. zero B. 10 N up or down C. 10 N to the right D. 10 N to the left

10 N to left

4) A man travels to a planet that has twice the radius of the earth and twice the mass. His weight on that planet compared to his weight on earth is: A) doubled. B) tripled. C) halved. D) cannot be determined with the information given.

C but this confused radius as distance Explanation: F = GM1M2/r2 (PHY 2.4), where G is the gravitational constant, M1 the mass of the planet, M2 the mass of the man and r the distance from the center of mass of the planet and the man. Therefore, F is proportional to M1/r2. If M1 is doubled, F will double. If r is doubled, F will be quartered. Thus, if M1 and r are both doubled, F is halved

Two planets triple the distance in between them. What will be the ratio of the new gravitational force to the old gravitational force between these two planets? A. 1/3rd B. 3 times C. 1/9th D. 9 times

C Correct choice: C Explanation: Any 2 masses experience gravitational forces whether they are chairs, computers, rocks, planets, people or asteroids. You are expected to know the inverse square relationship between force and distance. Clearly, if the distance increases, there will be less and less of a force between 2 masses. Newton's Law of Gravity tells us that . . F = G(m1m2/r2) So if you double r, you will get ¼ of the force; if you triple r, you get 1/9th of the force.

2) Suppose that an object is held at rest over a surface and dropped. The velocity on impact with the surface is 10 m/s. Neglecting air resistance, from approximately what height above the surface was the object dropped? (Gravitational acceleration = 9.8 m/s2) A) 5 m B) 10 m C) 19.6 m D) Cannot be determined without knowing the mass of the object.

Correct Answer: A Explanation: Any object relatively close to the surface of the Earth is drawn downward with a constant acceleration of 9.8 m/s2. Since the acceleration is constant, we can solve for the duration of the fall from: Vf = Vo + at 10 = 0 + (10)t t = 1 s (we have taken the vertically downward as the reference positive direction for velocity and acceleration; g is estimated at 10 because the answers are sufficiently far apart despite the information provided in the question). Next, we can solve for the length of the fall from: X = Vot + at2 X = 0 + 1/2(10)(1)2 = 5 m

6) A stone is dropped into a well. Ten seconds later a splash is heard. How deep is the water level below the well opening (ignoring the time taken for sound to travel)? A) 1000 m B) 500 m C) 50 m D) 200 m

Correct Answer: B Explanation: d = vinitialt + (1/2)at2 = (1/2)gt2 = 5 x 100 = 500 m

A student throws a rock at a 60° angle over a lake. The rock leaves his hand at a velocity of 20 m/s. How far away from the student will the rock shadow appear on the ground after 1 second, assuming the shadow projects straight down from the rock? A) 5 m B) 7 m C) 10 m D) 12 m

Correct Answer: C Explanation: Because the question asks for the distance between the rock's shadow and the student, the distance will be a horizontal distance. The horizontal velocity of the rock is given by vx= v•cosθ= v•cos60o= v (1/2). Thus vx=10 m/s. So, distance x is given by x= (vx)(t)= 10 m/s (1s)= 10 m.

5) A 10 kg block is accelerated at 2 m/s2 up a frictionless plane inclined at 30 degrees to the horizontal. The force acting parallel to the plane pushing the block upwards is: A) 100 N. B) 25 N. C) 50 N. D) 70 N.

Correct Answer: D Explanation: Weight of the block = 10 kg x 10 m/s2 = 100 N Force due to gravity pulling the block down the plane = sin(30) x 100 N = 50 N Net force accelerating the block = ma = 10 x 2 = 20 N Thus: external force applied to block parallel to the plane = 20 + 50 = 70 N

A 5 kg box is at rest on an inclined plane at 45 degrees to the horizontal. Friction is present, gravity can be estimated at 10 m/s2. The net force in newtons acting on the object is which of the following? A) 0 B) 25 C) 50 D) 75

Correct choice: A Explanation: An object that is in an inertial frame of reference (meaning it is either not moving or it is moving with constant velocity; i.e. it is NOT accelerating) must have no unbalanced force (i.e. no net force). Recall from Newton's Second Law that a net force must result in acceleration (F = ma).

Consider the Law of Gravitation where the force of gravity F is: F = G(m1m2/r2) Which of the following represents the gravitational constant G in the fundamental dimensions of mass (M), length (L) and time (T)? A. M-1L3T-2 B. M2L3T-2 C. M-1L-3T-2 D. M2L3T2

Correct choice: A Explanation: This is a classic example of dimensional analysis. First, you should know that the unit of force is a newton which is also a kg m/s2 (even if you did not memorize that, it is easy to work out since F = ma so the units of F must be kg multiplied by acceleration which is m/s2). Of course the 2 m's in the Law of Gravitation represent masses (M) and the r represents a distance or length (L). So we get: F = G(m1m2/r2) Now transferring to the fundamental quantities except for G: MLT-2 = G(M)2L-2 Divide both sides by (M)2L-2 and isolate for G: G = M-1L3T-2 ... which is the same as m3kg-1s-2.

1) Given the acceleration due to gravity of 32 ft/s2, what force is required to lift an object that weighs 16 lbs vertically with an acceleration of 16 ft/s2? A) 20 lb B) 24 lb C) 32 lb D) 36 lb

Correct choice: B Explanation: W = mg = 16 lb Thus: m = W/g = 16/32 = 1/2 slugs (no kidding!) ∑F = ma + mg = (1/2)16 + 16 = 24 lb --m = 1/2 a= 16 mg = Weight= 16 (note: the sum of forces is equal to the weight mg of the object plus the additional force ma causing its acceleration).

75 N force is used to push a large container weighing 400 N. The container does not move. What is the frictional force exerted by the floor on the container? A) Less than 75 N B) Exactly 75 N C) Between 75 N and 400 N D) Exactly 400 N

Correct choice: B Explanation: If the container is not moving, then the net force must be zero (a net force would result in acceleration; PHY 2.1). If a push of 75 N results in no movement then there must be a static frictional force equal and opposite to the push.

A horizontal force of 100 N pushes a 300 N container across a floor at a constant speed. What is the force of friction acting by the floor on the container? A. 50 N B. 100 N C. 150 N D. 300 N

Correct choice: B Explanation: If the speed is constant, then the forces are balanced, i.e. they add up to zero.100 N of kinetic friction (backwards, i.e. friction is always opposite to the direction of motion) is balancing the 100 N push (forward). Another way to think of it: Newton's 2nd Law can be summarized as a net force means acceleration. So, if there is constant speed, there is no acceleration, thus no net force. Do not be distracted by the weight of the object since weight (mg) is a force that acts straight downwards (i.e. towards the center of the Earth). Thus it acts perpendicular to the push and to friction and thus has no effect in the x-axis (example: push straight down on a table: will it move left or right? Neither.).

Which of the following is consistent with the SI Unit for frictional force and weight? A) J*s B) kg*m/s2 C) kg*m/s D) The pound

Correct choice: B Explanation: The frictional force and weight are both forces. Remember that force is mass (kg) times acceleration (m/s2). So a newton is kg*m/s 2. This is paying attention to units or dimensional analysis. Expect to use dimensional analysis on the real exam.

A 1 kg cart and a 2.0 kg cart are pushed apart by an expanding spring. If the average force on the 1.0 kg cart is 1.0 N, what is the average force on the 2.0 kg cart? A. 0.0 N B. 0.5 N C. 1.0 N D. 2.0 N

Correct choice: C Explanation: The situation may seem complicated but it comes down to Newton's third law which states that objects exert an equal but opposite force on each other. If you punch a wall with a force of 10 N (not recommended!) then the wall will exert a force of 10 N on your fist.

What is the approximate force on an electron, emitted from the cathode, which experiences an acceleration of 2.5 x 1012 m s-2? Mass of electron = 9.1 x 10-31 kg 1eV = 1.6 x 10-19J A. 2.8 x 10-6N B. 1.4 x 10-6N C. 2.3 x 10-18N D. 1.2 x 10-18N

Correct choice: C Explanation: Using Newton's Second Law, F = ma = (9.1 x 10-31 kg)(2.5 x 1012m s-2) = 9 x 25 x 10-31 x 1011 = 225 x 10-20 = 2.25 x 10-18 N (approx.) 1 eV = 1.6 x 10-19J is irrelevant.

An object is launched at 45 degrees from the horizontal. Neglecting air resistance, the final horizontal velocity would be equal to: A. (1/2) the initial horizontal velocity. B. (displacement)(sine 45). C. (displacement)(cosine 45) D. the initial horizontal velocity.

Correct choice: D Explanation: Since there is no air resistance, nothing opposes the horizontal velocity so it is constant (the gravity g points down), so there is no acceleration in the x direction, meaning that the initial velocity and the final velocity would be the same. Also, acceleration is defined as delta v/delta t (i.e. the change in velocity over the change in time; PHY 1.4.1) so if the acceleration is 0 then there can be no change in velocity over the time that the object is in the air.

A cannon is angled to the horizon such that, in the absence of gravity, its shells travel 4 feet horizontally for every 3 feet vertically. When a shell is fired, its speed on leaving the muzzle is 500 m/s. How far does the shell travel before hitting the ground (ignoring air resistance)? A. 20 km B. 18 km C. 12 km D. 24 km

Correct choice: D Explanation: A right triangle with sides 3, 4 and 5 units is formed when we take the muzzle velocity as the hypotenuse. Because it is a ratio (3:4:5), the actual units if the triangle are not relevant. Vertical component of muzzle velocity = 3/5 x 500 = 300 m/s Time to reach maximum height = vvertical/g = 300/10 = 30 s Time shell travels in the air = 2 x 30 = 60s Horizontal range = vhorizontal x t = 4/5 x 500 x 60 = 24,000 m Background check: If you did not instantly recognize the 3:4:5 triangle, you could have used the Pythagorean Theorem (PHY 1.1.1). Here is a colorful little reminder from high school math (ref: math.okstate.edu/geoset/Dictionary/3-4-5.htm).

If the sum of all the forces acting on a moving object is zero, the object will: A. slow down and stop. B. change the direction of its motion. C. accelerate uniformly. D. continue moving with constant velocity.

Correct choice: D Explanation: For something to change direction or accelerate there must be an unbalanced or net force acting on it (Newton's 2nd Law). Also, Newton's 1st Law states that an object in motion will stay in motion at constant velocity (a form of inertia; PHY 4.2) unless acted upon by an unbalanced force.

1) What is the SI unit for distance? A. kilometers B. meters C. centimeters D. feet

Meters

4) Approximately what magnitude a force would have to be applied in addition to the frictional force in order for the block in experiment 1 not to move on surface 3 when the plane is 60° to the horizontal? (note: sin 60 = 0.87; cos 60 = 0.5) A) 1.4 N B) 2.2 N C) 7.4 N D) 9.8 N CH 3

Net force being zero, and the force of friction being the force of static friction, as the block does not move, ΣF = Fgravity + Ffriction = mg sin θ + (μ∙N) = mg sin θ + μ∙-mg cos θ = (2 kg)(10 m/s2)(sin 60) - 1∙(2 kg)(10 m/s2)(cos 60) = (2 kg)(10 m/s2)(0.87) - 1∙(2 kg)(10 m/s2)(0.5) = (20 x 0.87) - (20 x 0.5) = 17.4 -10 = 7.4 N

How far will a brick starting from rest fall freely in 3.0 seconds? A. 22 m B. 44 m C. 67 m D. 88 m

Result: Your answer is correct. Your choice: B Correct choice: B Explanation: One of the equations of kinematics is d = vit + 1/2at2. The block starts from rest so the initial velocity is zero. The equation becomes d = 1/2at2. The only acceleration the brick feels is the acceleration due to gravity = 9.81m/s2 which we can estimate at 10. d = 1/2(10)(9) = 1/2(90) = 45 m

An object experiences 3 forces: (i) 8 N south; (ii) 11 N north; (iii) 4 N east. What is the magnitude of the resultant force? A. 3 N B. 5 N C. 7 N D. 23 N

Result: Your answer is incorrect. Your choice: N/A Correct choice: B 11 N north and 8 N south can be translated to 11 - 8 = 3 N in the north direction. With 4 N east (i.e. perpendicular, at right angles) we get the classic 3-4-5 triangle because of the Pythagorean theorem (PHY 1.1.1) where r is the resultant: r2 = x2 + y2 = (3)2 + (4)2 = 9 + 16 = 25 r = 5 N and since the resultant is drawn from the tail of the first vector to the tip of the last, we can see that the resultant vector direction is northeast. ( Vector is always drawn from tail to tip)

1) Consider a situation where you accidentally bump into a vase which then tilts briefly, but then returns to upright. When you initially bumped the vase, its center of gravity moved (gravitational potential energy is given by weight, mg, times height): A. upward and its gravitational potential energy increased. B. downward and its gravitational potential energy decreased. C. downward and its gravitational potential energy increased. D. upward and its gravitational potential energy decreased.

Your choice: C Correct choice: A Explanation: The vase is in a stable equilibrium, as evidenced by its tendency to return to that equilibrium after being bumped. To obtain this stable equilibrium, the vase's potential energy must increase as it tilts so that it accelerates in the direction that reduces its potential energy as quickly as possible--namely back toward equilibrium (i.e. the height h of its center of gravity, COG, increased making the potential energy higher before returning to a more stable position). Notice that the circular/spherical object in the diagram below is in neutral equilibrium and the center of gravity does not change. The center of gravity in unstable equilibrium is lowered making the unstable object more stable (image Ref. pt.ntu.edu.tw/hmchai/PTglossary).

Ignoring air resistance, the horizontal component of a projectile's acceleration: A. continuously decreases. B. continuously increases. C. is zero. D. remains constant but not necessarily zero.

Your choice: N/A Correct choice: C Explanation: Since there is no air resistance, nothing opposes the horizontal velocity (i.e. the object cannot slow down in the x direction) so it is constant (the gravity g points down), so there is no acceleration in the x direction.

2) On which surface would the block used in experiment I experience the greatest acceleration when the plane is inclined at the respective angle θk? A) Surface 2 B) Surface 3 C) Surface 4 D) The acceleration of the blocks remains constant for all surfaces. Ch 3

d Explanation: If the question had asked: On which surface would the block used in experiment I experience the greatest acceleration when the various planes are at 60 degrees? Then we would apply straightforward Physics: Friction acts against the direction of motion. So less friction means that there will be more acceleration down the plane (i.e. less resistance against the acceleration due to gravity). So among the answer choices, the lowest μk would result in the greatest acceleration, and so the answer would be A. Now we need to carefully reassess the meaning of the 3rd paragraph in the passage which begins with 'Researcher 1': "Once the block is moving, it will continue to move even if the plane makes a smaller angle with the horizontal. The smallest angle with which this occurs is also measured, θk (see Table 1)." We must reason that the object is moving at constant velocity. In other words, the acceleration due to gravity is counterbalanced by kinetic friction. The acceleration for all 3 surfaces at their respective angles θk must be zero, and thus constant.