HELP! 10 PTS! Two objects of different mass have equal, non-zero kinetic energies. Which object has the greater momentum?

1 answer:

4 0

The heavier one. The explanation for this is that <span>momentum is directly proportional to mass. The more mass there is, the more momentum.

This is as simply as I could explain it, hope I helped :) </span>

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Suppose the earth suddenly came to halt and ceased revolving around the sun. The gravitational force would then pull it directly

AleksAgata [21]

Answer:

613373.65233 m/s

Explanation:

M = Mass of Sun = $1.989\times 10^{30}\ kg$

m = Mass of Earth

v = Velocity of Earth

r = Distance between Earth and Sun = $147.12\times 10^{9}\ m$

$r_e$ = Radius of Earth = $6.371\times 10^6\ m$

$r_s$ = Radius of Sun = $695.51\times 10^6\ m$

In this system it is assumed that the potential and kinetic energies are conserved

$\dfrac{1}{2}Mv_2-\dfrac{GMm}{r_e+r_s}=0-\dfrac{GMm}{r}\\\Rightarrow v=\sqrt{2GM(\dfrac{1}{r_e+r_s}-\dfrac{1}{r})}\\\Rightarrow v=\sqrt{2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}(\dfrac{1}{6.371\times 10^6+695.51\times 10^6}-\dfrac{1}{147.12\times 10^{9}})}\\\Rightarrow v=613373.65233\ m/s$

The velocity of Earth would be 613373.65233 m/s

3 0

3 years ago

Which term describes an area of low density in a longitudinal wave?

LiRa [457]

Answer:

Explanation:

Low density is an area of rarefaction.

6 0

3 years ago

Read 2 more answers

I NEED HELP ASAP PLEASEEEE.A roller coaster ride starts with the roller coaster car being pulled to the top of the first hill wi

Sedaia [141]

I assuming they are not going to be very big just small enough for the car to get over the top and not go backwards. The coasting is all the momentum thy have to get over the hill and they don't have a lot of momentum. hope this helps ☺

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3 years ago

A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after bein

Elis [28]

Answer:

A. 49 joules

Step By Step Solution:

Via kinematic equations we can must first determine the velocity of this object after 10 meters (since halfway of 20 meters is 10).

$v_{final}^2=v_{initial}^2+2gd$

Where the initial velocity is zero since it's held at a height of 20 meters, hence it's not moving initially, g is the average gravity on Earth 9.8 m/s^2 and d is the distance this object will travel so 10 meters. By plugging these values in we obtain:

$v_{final}^2=0+2(9.8)(10)\\\\v_{final}=\sqrt{2(9.8)(10)}=14m/s$