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Crank
3 years ago
12

A boxcar at a rail yard is set into motion at the top of a hump. The car rolls down quietly and without friction onto a straight

, level track where it couples with a flatcar of smaller mass, originally at rest, so that the two cars then roll together without friction. Consider the two cars as a system from the moment of release of the boxcar until both are rolling together. (a) is the mechanical energy of the system conserved? (b) is the momentum of this system conserved? Next, consider only the process of the boxcar gaining speed as it rolls down the hump. Consider the boxcar and the Earth as a system. (a) is the mechanical energy of the system conserved? (b) is the momentum of this system conserved? Finally, consider the two cars as a system as the boxcar is slowing down in the coupling process. (a) is the mechanical energy of the system conserved? (b) is the momentum of this system conserved?

Physics
1 answer:
enyata [817]3 years ago
3 0

Answer:

I have answer with explanations in each of the six (6) special cases.

Explanation:

Please find attached the detailed solution to all your questions.

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Romashka [77]

Answer:

answer is 18.58 because

5 0
3 years ago
Read 2 more answers
A Jack Rabbit hops 12.8 meters per second. How long would it take for him to hop 353 m? Round to the nearest tenth place. ​
adoni [48]

Time = Distance/speed

353/12.8 =27.57 seconds

4 0
3 years ago
An astronaut exploring a distant solar system lands on an unnamed planet with a radius of 2530 km. When the astronaut jumps upwa
Natali [406]

Answer:

1.38*10^18 kg

Explanation:

According to the Newton's law of universal gravitation:

F=G*\frac{m_a*m_p}{r^2}

where:

G= Gravitational constant (6.674×10−11 N · (m/kg)2)

ma= mass of the astronaut

mp= mass of the planet

F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2

so:

m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg

7 0
3 years ago
A ball is dropped from a height of 20 meters. At what height does the ball have a velocity of 10 meters/second?
borishaifa [10]

Answer:B

Explanation:

Initial velocity, u=0m/s

Distance,s=20m

a=+g=9.8m/s*s

Using v*v=u*u+2gs

v*v=0+2*9.8*20

v*v=392

v=19.8

When s=20m, v = 19.8m/s

Therefore when v = 10m/s, s= 10*20/19.8

s =10.1m

6 0
3 years ago
14. Which one of the following pictures shows the object that is the most dense? *
rewona [7]

Answer:

B

Explanation:

Density is about how closely compact molecules are. (^-^)

5 0
2 years ago
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