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marissa [1.9K]
3 years ago
15

Explain how energy is transferred in an impact situation such as a car crash.

Physics
1 answer:
DaniilM [7]3 years ago
8 0
During a car crash, energy is transferred from the vehicle to whatever it hits, be it another vehicle or a stationary object. ... The object that was struck will either absorb the energy thrust upon it or possibly transfer that energy back to the vehicle that struck it.

I HOPE THIS HELPSS???
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In each cycle, a heat engine an input of 1940 J of heat and exhausts 1480 J of heat. What is the thermal efficiency?
AleksAgata [21]

Answer:

0.237 (23.7 %)

Explanation:

The thermal efficiency of an engine is given by:

\eta=\frac{W}{Q_{in}}

where

W is the useful work output of the engine

Q_{in} is the heat in input

Here we have:

Q_{in}=1940 J

and the work done is the total heat in input minus the heat exhausted:

W=1940 J - 1480 J=460 J

So, the efficiency is

\eta=\frac{460 J}{1940 J}=0.237 (23.7 %)

8 0
3 years ago
(a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you ca
Colt1911 [192]

a) 52.5 m

b) 16.0 m/s

Explanation:

a)

The motion of a car slowed down by friction is a uniformly accelerated motion, so we can use the following suvat equation:

v^2-u^2=2as

where

v = 0 is the final velocity (the car comes to a stop)

u = 28.7 m/s is the initial velocity of the car

a is the acceleration

s is the stopping distance

For a car acted upon the force of friction, the acceleration is given by the ratio between the force of friction and the mass of the car, so:

a=\frac{-\mu mg}{m}=-\mu g

where:

\mu=0.80 is the coefficient of friction

g=9.8 m/s^2 is the acceleration due to gravity

Substituting and solving for s, we find:

s=\frac{v^2-u^2}{-2\mu g}=\frac{0-(28.7)^2}{-2(0.80)(9.8)}=52.5 m

b)

In this case, the car is moving on a wet road. Therefore, the coefficient of kinetic friction is

\mu=0.25

Here we want the stopping distance of the car to remain the same as part a), so

s=52.5 m

We can use again the same suvat equation:

v^2-u^2=2as

And since the final velocity is zero

u = 0

We can find the initial velocity of the car:

v=\sqrt{2as}=\sqrt{2\mu gs}=\sqrt{2(0.25)(9.8)(52.5)}=16.0 m/s

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The answer is B

Explanation:

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