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KatRina [158]
3 years ago
10

A car mass 600kg starts from rest moving uniform acceleration 0.2 m/s^2 after 60 seconds collides with stationary pick up van of

mass 400kg gets locked with it and moves together with velocity 7.2 m/s
a) DOES THE PHENOMENON OF THE STEM SUPPORT THE PRINCIPLE OF CONSERVATION OF MOMENTUM? GIVE YOUR OPINION BY MATHEMATICAL ANALYSIS.​
Physics
1 answer:
galben [10]3 years ago
5 0

Answer:

the phenomenon of the system supports the principle of conservation of momentum.

Explanation:

The law of conservation of momentum says that:

Initial Momentum = Final Momentum

So, first we calculate initial momentum of the system:

Initial Momentum  = m₁u₁ + m₂u₂

where,

m₁ = mass of car = 600 kg

m₂ = mass of van = 400 kg

u₁ = Initial Speed of Car

For initial speed of car, we use:

Vf = Vi + at

Vf = 0 m/s + (0.2 m/s²)(60 s)

Vf = u₁ = 12 m/s

u₂ = Initial Speed of Van = 0 m/s

Therefore,

Initial Momentum  = (600 kg)(12 m/s) + (400 kg)(0 m/s)

Initial Momentum  = 7200 Ns   --------------- equation (1)

Now, for the final momentum:

Final Momentum  = m₁v₁ + m₂v₂

where,

v₁ = v₂ = Final Speed of Car and van (both are locked) = 7.2 m/s

Therefore,

Final Momentum = (600 kg)(7.2 m/s) + (400 kg)(7.2 m/s)

Final Momentum = 7200 Ns   ------------- equation (2)

Comparing equation (1) and (2):

Initial momentum = Final Momentum

<u>Hence, the phenomenon of the system supports the principle of conservation of momentum.</u>

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7 0
3 years ago
abody starts from rest and accelerate uniformly at 8m/s/s.calculate the distance travelled by the body in 10s? ​
vagabundo [1.1K]

Explanation:

using the formula: S=ut+½gt², where u=0, S=?, g=8m/s², t=10seconds.

S=ut+½gt² ("ut" term will cancel because u=0).

=> S= ½gt²

=>S = ½×8×10²

=>S = 4×100

=>S = 400m .

Therefore, the distance traveled by the body in 10s is 400m.

hope this helps you.

6 0
2 years ago
A velocity-time graph shows how what changes over time.
Mnenie [13.5K]

Answer:

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6 0
2 years ago
A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s
Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

T=m\frac{v^2}{r}

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N

5 0
3 years ago
If we drop different weights from same height which will fall first to ground? Both of them hit the ground at the same time righ
crimeas [40]
Yes tthey would both hit the ground
5 0
2 years ago
Read 2 more answers
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