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2 years ago

What will happen if your brakes stop your wheels before your car's kinetic energy has been absorbed?

2 answers:

8 0

The friction between your wheels and the brakes, along with the friction between your tires and the road, causes your car to come to a stop. ... If this happens, your brakes will stop your wheels before your car's kinetic energy has been absorbed. When this happens, your car will keep moving, and your wheels will skid

iris [78.8K]2 years ago

8 0

Answer:Explained

Explanation:

If brakes stops wheels before car's kinetic energy has been absorbed then car will still be in motion until its kinetic energy will be zero and while kinetic is been absorbed the will skid on the road because there is still the kinetic energy in car which is being gradually absorbed by friction.

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3 years ago

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A particle moving on a plane curve what is the degree of freedom

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3 years ago

A plane flies from base camp to lake a, 200 km away in the direction 20.0° north of east. after dropping off supplies it flies t

Liono4ka [1.6K]

Distance of lake a is 200 km at 20 degree north of east

distance between lake a and b is 230 km at 30 degree west of north

now the distance between base and lake b is given as

$d = d_1 + d_2$

given that

$d_1 = 200 cos20 i + 200 sin20 j$

$d_1 = 187.94 i + 68.4 j$

$d_2 = -230 sin30 i + 230 cos30 j$

$d_2 = -115 i + 199.2 j$

now the total distance is

$d = (187.94 - 115)i + (199.2 + 68.4)j$

$d = 72.94 i + 267.6 j$

now the magnitude of the distance is given as

$d = \sqrt{72.94^2 + 267.6^2}$

$d = 277.4$

also the direction is given as

$\theta = tan^{-1}\frac{267.6}{72.94}$

$\theta = 74.7 degree$

<em>so it is 277.4 km at 74.7 degree North of East</em>

8 0

3 years ago

It decomposes in a first-order reaction with a rate constant of 14 s–1. how long would it take for an initial concentration of 0

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The rate constant of a reaction can be computed by the ratio of the changes in the concentration and time take taken for it to decompose. Thus, if the rate constant is given to be 14 M/s, we have

$rate = \frac{-(C_{new} - C_{old})}{t}$

where C are the concentration values and t is the time taken for it to decompose.

$14 = \frac{-(0.02 - 0.06)}{t}$
$t = 0.003 s$

Thus, it will take 0.003 s for it to decompose.
Answer: 0.003 s

4 0

3 years ago