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GenaCL600 [577]
3 years ago
5

A person holding a 15.0 kg containing one 50.0 g bullet is riding on a train that is traveling at 75.0 km/h east. If the man fir

es the gun and the bullet moves with a velocity of 350 m/s east(relative to the train), what is the velocity of the gun relative to the ground
Physics
1 answer:
Lana71 [14]3 years ago
8 0

Answer:

The velocity of the gun relative to the ground is 19.66 m/s

Explanation:

Given data,

The mass of the gun, M = 15.0 kg

The mass of the bullet, m = 50 g

The velocity of the train, v = 75 km/h

                                           = 20.83 m/s

The velocity of bullet relative to train, V' = 350 m/s

The velocity of bullet relative to ground, V = 350 + 20

                                                                       = 370 m/s

According to the law of conservation of momentum,

                                Mv' + mV' = 0

                                   v' = -\frac{mV'}{M}

                                   v' = -\frac{0.050\times 350}{15}

                                      = -1.17 m/s

Therefore, the velocity of the gun with,

                                   v₀ = V + v'

                                        = 20.83 - 1.17

                                         = 19.66 m/s

Hence, the velocity of the gun relative to the ground is 19.66 m/s

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A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.

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y(1s) - y(1/2s) =  - 3.675 m  

The cat falls 3.675 m between time 1/2 s and 1 s.

Explanation:

Given data

time=1/2 sec to 1 sec

v(t)=-9.8t m/s

To find

Distance

Solution

As the acceleration as first derivative of velocity with respect to time  

So

acceleration(-g)=  dv/dt

Solve it

dv  =  a dt

dv =  -g dt

v - v₀  =  -gt

v=  dy/dt

dy  =  v dt

dy =  ( v₀ - gt ) dt

y(1s) - y(1/2s)  =  ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]

y(1s) - y(1/2s)  = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]

y1s - y1/2s  = ( - 4.9 m/s² ) ( 3/4 s² )

y(1s) - y(1/2s) =  - 3.675 m  

The cat falls 3.675 m between time 1/2 s and 1 s.

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Explanation:

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<span>The original area was 94.71 * 189.92 = 17939.9 cm^2 </span>
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