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Kisachek [45]
4 years ago
12

During a demolition derby, vehicle 1 collided vehicle 2 going in the opposite direction and came to an immediate stop. If the ve

hicles are of different masses and were not travelling at the same velocity, what can be concluded about them? *
Physics
1 answer:
snow_tiger [21]4 years ago
7 0

Answer:

Their change in momentum is the same in magnitude and opposite in direction

Explanation:

The momentum of an object is defined as:

p=mv

where

m is the mass of the object

v is the velocity of the object

Therefore, the change in momentum of an object is

\Delta p = m\Delta v

where \Delta v is the change in velocity.

During a collision, the force experienced by an object is equal to the rate of change of momentum:

F=\frac{\Delta p}{\Delta t}

where \Delta t is the duration of the collision.

According to Newton's third law of motion, the force exerted by vehicle 1 on vehicle 2 during the collision is equal (and opposite) to the force exerted by vehicle 2 on vehicle 1, so

F_1=-F_2

Which means

\frac{\Delta p_1}{\Delta t}=-\frac{\Delta p_2}{\Delta t}

And since the duration of the collision is the same for the two vehicles, this becomes

\Delta p_1 =-\Delta p_2

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4 years ago
A proton and an electron in a hydrogen atom are separated on the average by about 5.3 × 10−11 m. What is the magnitude and direc
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Answer:

1. 5.12068 × 1011 N/C away from the proton

Explanation:

The electric field produced by a single point charge is given by:

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q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

q=1.6\cdot 10^{-19}C is the charge of the proton

r=5.3\cdot 10^{-11} m is the distance at which we want to calculate the field

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Substituting into the formula,

E=(8.99\cdot 10^9 Nm^2C^{-2})\frac{1.6\cdot 10^{-19}C}{(5.3\cdot 10^{-11}m)^2}=5.12068\cdot 10^{11} N/C

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1. 5.12068 × 1011 N/C away from the proton

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Answer:

góc phản xạ 15°

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