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marshall27 [118]
3 years ago
5

In elastic collisions,total kinetic energy and momentum are both conserved. On a frictionless track,cart 1 is moving with a cons

tant,rightward(+) velocity of 1.0m/s.Cart 2 is also moving rightward with constant velocity of 5.0m/s. After a while,cart 2 collides cart 1 from behind.(It is an elactic collision.)If the final velocity of cart 2 becomes 3.0m/s, what is the final velocity of cart 1?(SHOW ALL YOUR WORK.)
Physics
1 answer:
kari74 [83]3 years ago
5 0

Answer:

Explanation:

mass of cart 1, m1 = m

mass of cart 2, m2 = m

initial velocity of cart 1, u1 = 1 m/s

initial velocity of cart 2, u2 = 5 m/s

final velocity of cart 2, v2 = 3 m/s  

Let the final velocity of cart 1 is v1.

Use the conservation of momentum

m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2

m x 1 + m x 5 = m x v1 + m x 3

6 = v1 + 3

v1 = 3 m/s

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An object is 39 cm away from a concave mirrors surface along the principles axis. If the mirrors focal length is 9.50 cm, how fa
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Answer:

12.6 cm

Explanation:

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5 0
3 years ago
Please help answer question​
nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

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R = W

or

\dfrac{1}{2}\rho CAv_t^2=mg

Where

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C is drag coefficient

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C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01

So, the drag coefficient is 1.01.

4 0
2 years ago
A rabbit is hopping along at an approximately constant speed of 3.9 m/s. The rabbit passes a crouched cat ready to chase the rab
Firlakuza [10]

Answer:

t  = 7,8 s

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From the instant, the rabbit passes the cat. The cat star running acceleration of 0,5 m/s² .

When the cat arrives at the speed of 3,9 m/s the cat catches the rabbit

Then for the cat arrives at 3,9 m/s nedds

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3,9 ( m/s) = 0,5 ( m/s² ) * t

t  = 7,8 s

v  =  3,9 m/s =

4 0
2 years ago
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