Physics 1050 Midterm

7.456452

4. A woman is planning on running a 10K (10 km) race a. How many yards will she race? (1 inch = 2.54 cm) 10936.1 b. What is the distance of this race in miles? (1 mile = 5280 feet) 6.21371 c. If she completes the race in 50 minutes, what is her average pace in miles per hour? 7.456452 a. you will set up a proportion knowing that 1 yard = 0.0009144 kilometers b. you will set up a proportion knowing that 1 mile = 1.60934 kilometers c. Using the answer, you got for b, you will set up as the picture shows and divide for the correct answer

answer says (-) but both + and - worked??) 45000

1. A car crashes into a wall at 60 m/s and is brought to rest in 0.1 s. Calculate the average force exerted on a 75 kg test dummy by the seat belt. Answer: 45000N Work: (-60)/0.1= -600 75kg*-600= -45000N (The answer says (-) but both + and - worked??)

Answer: 16 times the KE

1. A car moves 4 times as fast as another identical car. Compared to the slower car, the faster car has Work: K.E. (car 1) = ½x1000x1/2 K.E. = 500 joules the second car goes 4 times faster so K.E. (car 2)= 1/2x1000x4^2 KE (car 2) = 1/2x1000x16 = 8000 joules so 8000 is 16 times bigger than 500, so if something has the same mass but travels 4 times faster it will always have 16 times more kinetic energy Answer: 16 times the KE

2.1m/s2

1. A car takes 14 s to go from v = 0 to v = 30 m/s at approximately constant acceleration. If you wish to find the distance traveled using the equation d = 1/2 at2, what value should you use for a? 2.1m/s2 Because we know a is our acceleration, we can easily find this because acceleration is the change in velocity over time. Since our velocity went from 0 m/s to 30 m/s, the change in velocity is 30. We divide 30 by 14 because that is the time it took to get to 30 m/s. Therefore, we get the answer 2.1m/s2 30 / 14 = 2.1 m/s2 Divide the second number by the first number.

Answer: twice as much work

1. If you push an object twice as far while applying the same force, you do Answer: twice as much work

Answer: 8 km/hr

1. What is the impact speed when a car moving at 80 km/h bumps into the rear of another car traveling in the same direction? Answer: 8 km/hr Work: 80-72= 8

Answer: 9.8 m/s

10. In the absence of air resistance, by how much does it increase each second while descending? Answer: 9.8 m/s

**but opposite amounts of momentum. amounts of kinetic energy.

10. When a rifle is fired, it recoils so both the bullet and rifle are set in motion. The rifle and bullet ideally acquire equal **but opposite amounts of momentum. amounts of kinetic energy. both of these. none of these. (You guessed it..idk)

**1 m/s.

11. A rifle of mass 2 kg is suspended by strings. The rifle fires a bullet of mass 1/100 kg at a speed of 200 m/s. The recoil velocity of the rifle is about 0.01 m/s. none of these. 0.1 m/s. **1 m/s. 0.001 m/s. Work: m1 = 2 kg v1 = ? m2 = 0.01 kg v2 = 200 m/s m1v1 = m2v2 (2 kg) * v1 = (0.01 kg) * (200 m/s) (2 kg) * v1 = (2 kg-m/s) v1 = (2 kg-m/s) / (2 kg) v1 = 1 m/s

Answer: the times will be the same.

11. How much time is required for rising compared to falling? Answer: the times will be the same.

Work: 2 (2.998 x 10^8 m/s) year just google what is 2 light years Answer: 1.8921E+16

12. A "light-year" is the distance light (speed = 2.998 ✕ 108 m/s) travels in one year. How many meters are there in 2.00 light-years? Work: 2 (2.998 x 10^8 m/s) year just google what is 2 light years Answer: 1.8921E+16

**0.83 m/s.

12. A 30-kg girl and a 25-kg boy face each other on friction-free roller blades. The girl pushes the boy, who moves away at a speed of 1.0 m/s. The girl's speed is **0.83 m/s. 0.45 m/s. 1.2 m/s. 0.55 m/s. Work: m1v1=m2v2 25*1=30*v2 25/30=v2 v2=0.83m/s

**6 m/s

13. A 5000-kg freight car runs into a 10,000-kg freight car at rest. They couple upon collision and move with a speed of 2 m/s. What was the initial speed of the 5000-kg car? 4 m/s none of these. 8 m/s 5 m/s **6 m/s Work: m1v1+m2v2=(m1+m2)v m1=mass(kg) of object one v1= velocity of object one in m/s m2= mass of object two v2= velocity of object two v= final velocity (5000kg)(v1)+(10000kg)(0 m/s)=(5000kg)(10000kg)(2 m/s) (5000kg)(v1)+0=30000 kgm/s v1= 6 m/s

Work: 1.8921E+16/1.496x10^11 google how many ay are there in 2 light year

13. An astronomical unit (AU) is the average distance from the Sun to Earth, 1.50 ✕ 108 km. How many AU are there in 2.00 light-year? Work: 1.8921E+16/1.496x10^11 google how many ay are there in 2 light year Answer:

Answer: 0.5 m/s

14. A 5 kg fish swimming 1 m/s (take this as the positive direction) swallows an absent minded 1 kg fish swimming toward it at a speed of 2 m/s. What is the speed v of the larger fish after lunch? Work: (5 x 1) + (1 x -2) = (2 + 1)v then 5 - 2 = 6v then v = 3/6 = .5 m/s Answer: 0.5 m/s

**The craft moves to the right.

14. An ice sailcraft is stalled on a frozen lake on a windless day. The skipper sets up a fan as shown. If all the wind bounces backward from the sail, will the craft be set in motion? If so, in what direction? The craft moves to the left. The craft does not move. **The craft moves to the right.

Answer: 7 m/s

15. A 5 fish swimming 1 m/s swallows an absent minded 1 kg fish swimming toward it at a velocity that brings both fish to a halt immediately after lunch. What is the velocity v of the smaller fish before lunch? Work: m1 x v1 + m2 x v2 = mf x vf then 5 x 1 + 1 x v2 = 6 x 0 Then 5 + v2 = 0 then v2 = -5 m/s that is 5 m/s toward the big fish. Answer: 7 m/s

Answer: 40

2. A TV set is pushed a distance of 2 m with a force of 20 N that is in the same direction as the set moves. How much work is done on the set? Work: 1°/ Work = Force x distance; so work is 20 * 2 = 40N Answer: 40

5.4 m/s2

2. A car takes 14 s to go from v = 0 to v = 75 m/s at approximately constant acceleration. If you wish to find the distance traveled using the equation d = 1/2 at2, what value should you use for a? 5.4 m/s2 See the explanation for problem 1. 75 / 15 = 5.4m/s2 Divide the second number by the first number.

Answer: 1.017

2. Lillian (mass 44.0 kg) standing on slippery ice catches her leaping dog (mass 15 kg) moving horizontally at 4.0 m/s. What is the speed of Lillian and her dog after the catch? Answer: 1.017 Work: v1 = 0 (Lillian's original speed) v2 = 4 (dog's initial speed) v3 = Lillian + dog = ? m1 = mass dog = 15 m2 = mass Lillian = 44 m2*v1 + m1*v2 = (m1 + m2) v3 44*0+15*4=(15+44)? 60=59? ?=1.017

Answer: 117

2. This question is typical on some driver's license exams: A car moving at 50 km/h skids 13 m with locked brakes. How far will the car skid with locked brakes at 150 km/h? Work: 13/50 x x/150 so do 13x150=1950 then do 1950/50=39 then do 39x3=117 Answer: 117

Answer: 10.4

2. What is the efficiency of the body when a cyclist expends 1200 W of power to deliver mechanical energy to her bicycle at the rate of 125 W? Work: 125/1200=.10416 move decimal for percent so you get 10.4 Answer: 10.4

Answer: 100 J.

3. A person on the edge of a roof throws a ball downward. It strikes the ground with 100 J of kinetic energy. The person throws another identical ball upward with the same initial speed, and this too falls to the ground. Neglecting air resistance, the second ball hits the ground with a kinetic energy of Answer: 100 J.

Answer: 700000m/s

3. Comic-strip hero Superman meets an asteroid in outer space, and hurls it at 700 m/s, as fast as a bullet. The asteroid is a thousand times more massive than Superman. In the strip, Superman is seen at rest after the throw. Taking physics into account, what would be his recoil velocity? Answer: 700000m/s Work: momentum=mass*velocity 700000=700*1000

Answer: heat

3. When a car is braked to a stop, its kinetic energy is transformed to TV set is pushed a distance of 2 m with a force of 20 N that is in the same direction as the set moves. How much work is done on the set? Work: google this b Answer: heat

Answer: twice as much

4. A 1000-kg car and a 2000-kg car are hoisted the same distance in a gas station. Raising the more massive car requires Work: the force required to raise the car at constant speed is equivalent to the weight of the car. Since the 2000 kg car weighs twice as much as the 1000 kg car it would require twice as much work to lift it the same distance. Answer: twice as much

Answer: 10.5 m/s

4. A 3.5 kg fish swimming 3 m/s swallows an absent minded 1 kg fish swimming toward it at a velocity that brings both fish to a halt immediately after lunch. What is the velocity v of the smaller fish before lunch? Answer: 10.5 m/s Work: momentum = mass x velocity (p=mv) Call the big fish - fish #1, and the little fish - fish #2. Now solve for v2, the velocity of the little fish. (m1v1)/m2 = v2 (3.5*3)/1=10.5

Answer: 1380

4. A car with a mass of 1200 kg moves at 23 m/s. What braking force is needed to bring the car to a halt in 20 s? Work: -23/20 = -1.15 then do force = mass x acceleration so do 1200x-1.15=-1380 I took off - sign Answer: 1380

1.5 x 1011 m

6. The Sun, on average, is 93 million miles from the Earth. Express how many meters this is using powers of ten notation. (Give a value between 1 and 10 and the correct exponent.) 1.5 x 1011 m Google how "how many meters is 93 million miles." For the answer, you must round to the nearest tenth and use the scientific notation given. See picture below.

Answer: 5.238 km/h

7. A railroad engine weighs three times as much as a freight car. If the engine coasts at 7 km per hour into a freight car that is initially at rest, how fast do the two coast after they couple together? Answer: 5.238 km/h Work: Mu1 + mu2 = (M + m)v as M = 3m in this example 3mu1 + mu2 = (3m + m)v 3m(1.94) + m(0) = (4m)v ---1.94=km/hr converted to m/s v = 3m(1.94) / 4m v = 3(1.94) / 4 v = 1.455 m/s v = 1.455 m/s = 5.238km/hr

Answer: 9.8

7. Suppose that a freely falling object were somehow equipped with a speedometer. By how much would its speed reading increase with each second of fall? Answer: 9.8

Answer: the middle (concave) hill

8. On which of these hills does the ball roll down with increasing speed and decreasing acceleration along the path? (Use this example if you wish to explain to someone the difference between speed and acceleration.) Answer: the middle (concave) hill

**a pickup truck speeding along a highway

8. Which of the following has the largest momentum relative to the Earth? the Science building on campus a Mack truck parked in a parking lot a dog running down the street a tightrope walker crossing Niagara Falls **a pickup truck speeding along a highway (Idk how to work this lol)

**rifle has more mass than the bullet. momentum is mainly concentrated in the bullet. force against the rifle is smaller than against the bullet.

9. A rifle recoils from firing a bullet. The speed of the rifle's recoil is small because the momentum of the rifle is smaller. **rifle has more mass than the bullet. momentum is mainly concentrated in the bullet. force against the rifle is smaller than against the bullet. (Again, idk...)

9 km/h

9. What is the impact speed when a car moving at 90 km/h bumps into the rear of another car traveling in the same direction at 81 km/h? 9 km/h Subtract the smaller number from the bigger number.

Answer: 9.8 m/s

9. When a ball player throws a ball straight up, by how much does the speed of the ball decrease each second while ascending? Answer: 9.8 m/s

125 m 10 s

A ball is thrown straight up with an initial speed of 50 m/s. (Neglect air resistance.) How high does it go? 125 m How long is it in the air? 10 s (2t) time velocity distance fallen ( seconds) ( m/s ) ( meters ) 0 0 0 1 10 5 2 20 20 3 30 45 4 40 80 5 50 125 6 60 180 7 70 245 8 80 320 9 90 405 10 100 500 t (10) t (1/2) (10) t2

0 m/s

A ball is thrown with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball when it gets to its highest point? 0 m/s (b) What is its velocity 2s before it reaches its highest point? 20 m/s (c) What is the change in its velocity during this 2s interval? 20 m/s (d) What is its velocity 2 s after it reaches its highest point? 20 m/s (e) What is the change in velocity during this 2 s interval? 20 m/s (f) What is the change in velocity during the 4 s interval? (Careful!) 40 m/s (g) What is the acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity? -10 m/s2

10936.1

A woman is planning on running a 10K (10 km) race. a. How many yards will she race? 10936.1 b. What is the distance of this race in miles? 6.21371 c. If she completes the race in 61.8 minutes, what is her average pace in miles per hour? 6.032728155 For a and b, see problem number 4 (it is exact). For c, using the answer, you got for b, you will set up as the picture shows and divide for the correct answer.

both?

Someone standing at the edge of a cliff (as in the figure) throws a ball nearly straight up at a certain speed and another ball nearly straight down with the same initial speed. If air resistance is negligible, which ball will have the greater speed when it strikes the ground below? The ball thrown downward will strike at a greater speed. The ball thrown upward will strike at a greater speed. Both will strike the ground below at the same speed.

10

Suppose that a freely falling object were somehow equipped with a speedometer. By how much would its speed reading increase with each second of fall? 10 m/s

.7694 s

Very few athletes can jump more than 2.4 ft (0.74 m) straight up. Use d = 1/2 gt2 and solve for the time one spends moving upward in a 2.4 foot vertical jump. Then double it for the "hang-time"--the time one's feet are off the ground. .7694 s

Answer: 68.6 m/s

What is the average velocity during this interval? Work: Average Velocity = (initial velocity+ final velocity)/2 = (0+137.2 m/s)/2 = 68.6 Answer: 68.6 m/s

130 m/s (13 x 10)

What is the instantaneous velocity of a freely falling object 13 s after it is released from a position of rest? 130 m/s (13 x 10) What is its average velocity during this interval? 65 m/s (13 x 5) How far will it fall during this time? 845 m (13 x 65 (answer from part b))

Answer: 137.2 m/s or -140 (questions were different for me and molly?)

What is the instantaneous velocity of a freely falling object 14 s after it is released from a position of rest? Work: 9.8x14= 137.2 m/s Answer: 137.2 m/s or -140 (questions were different for me and molly?)

10 m/s

When a ball player throws a ball straight up, by how much does the speed of the ball decrease each second while ascending? 10 m/s In the absence of air resistance, by how much does it increase each second while descending? 10 m/s

Answer: 5.10 m

1. A ball is thrown straight up with an initial speed of 10 m/s. (Neglect air resistance.) a) How high does it go? Work: 10^2=100 9.8x2=19.6 so do 100/19.6=5.10 m Answer: 5.10 m b) How long is it in air? Work: 10/9.8=1.02 then double it 1.02x2= 2.04 s Answer: 2.04 s

Answer: 960.4

. How far will it fall during this time? Work: Distance traveled (fall) = 68.6 m/s x 14 sec = 960.4 Answer: 960.4

7.569e+16 m

5. A "light-year" is the distance light (speed = 2.998 ✕ 108 m/s) travels in one year. a. How many meters are there in 8.00 light-years? 7.569e+16 m b. An astronomical unit (AU) is the average distance from the Sun to Earth, 1.50 ✕ 108 km. How many AU are there in 8.00 light-year? 505929 AU c. What is the speed of light in AU/h? 7.21436 AU/h I honestly have no earthly idea what this is, but google converts them. Copy and paste the parts a-c (one at a time). The initial question is irrelevant in our case. However, for part b, in order to get the correct answer, you must only copy and paste the question at the end. The first sentence does not matter.

Answer: 1.67m/s

5. A 2 kg fish swimming 3 m/s (take this as the positive direction) swallows an absent minded 1 kg fish swimming toward it at a speed of 1 m/s. What is the speed v of the larger fish after lunch? Answer: 1.67m/s Work: momentum M1V1 + M2V2 = (M1 + M2)Vf where M1 = 2 kg. V1 = 3 m/s M2 = 1 kg. V2 = 1 m/s Vf = final speed of the large fish = final speed of the small fish (2)(3)+(1)(-1)=(2+1)Vf ------the (-) is there because it's the velocity of the small fish Vf= 5/3 Vf= 1.67m/s

Answer: 6.25 m/s^2

5. A car takes 12 s to go from v = 0 to v = 75 m/s at approximately constant acceleration. If you wish to find the distance traveled using the equation d = 1/2 at2, what value should you use for a? Work: the average accelerations = (change of velocity)/(time elapsed). So (75m/s - 0m/s)/12s = 6.25 m/s^2 Answer: 6.25 m/s^2

Answer: Twice as much work

5. When an object is lifted 10 m, it gains a certain amount of potential energy. If the same object is lifted 20 m, its potential energy gain is Work: since 20 m is twice as much than 10 m the potential energy gain is twice as much. Answer: Twice as much work

Answer: 0.61m/s

6. A 4.5 kg little fish stops to tie his shoe lace just before meeting a hungry 7kg large fish swimming at 1 m/s. What is the speed v of the larger fish after lunch? Answer: 0.61m/s Work: (7 kg * 1 m/s) + (4.5 kg * 0 m/s) = 11.5 kg * ? m/s -----the 11.5 comes from the mass of small fish and large fish combined 7=11.5kg(?m/s) ?=0.61

Answer: 0.85 s

6. Surprisingly, very few athletes can jump more than 2.9 ft (0.88 m) straight up. Use d = 1/2 gt2 and solve for the time one spends moving upward in a 2.9 foot vertical jump. Then double it for the "hang-time" -- the time one's feet are off the ground. Work: d= ½ gt^2 so d=2.9 foot, g=32 plug in numbers to solve for T then double it for hang time. ½ 32t^2= t=Square root of 2d/g square root of 2(2.9 ft)/32 = .425 double it .425x2= 0.85 Answer: 0.85 s