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goblinko [34]
2 years ago
10

At a race car driving event, a staff member notices that the skid marks left by the race car are

Physics
1 answer:
krok68 [10]2 years ago
5 0

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).

We can calculate the initial velocity of the car (u) using the following kinematic equation.

v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

Learn more: brainly.com/question/14851168

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A 1,120 kg car is travelling with a speed of 40 m/s. Find its energy.
serious [3.7K]

Answer:

this vehichle has 896,000 jules of energy

Explanation: KE=1/2mv squared, KE is Kinetic energy. m is mass and v is velocity

8 0
3 years ago
5. Which pair of words, in order, correctly completes
sattari [20]

Answer:

B is an answer.

3 0
3 years ago
How many meters are in 32 kilometers
lukranit [14]

Answer:

32,000 m

Explanation:

Conversion

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5 0
3 years ago
If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object paral
leonid [27]

Answer:

\theta = tan^{-1}\mu

Explanation:

As we know that if the object is placed on the inclined plane then the force of friction on the object is counterbalanced by the component of the weight of the object along the inclined plane.

So we can say

F_f = mgsin\theta

now if we increase the inclination of the plane then the component of the weight weight along the inclined plane will increase and hence the friction force will also increase.

As we know that the limiting value or the maximum value of friction force at the static condition is given by

F_f = \mu N

N = mg cos\theta

so we have

F_f = \mu (mg cos\theta)

so we will have

mg sin\theta = \mu (mg cos\theta)

so now we have

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3 0
3 years ago
I will Give Brainliest To whoever actually answers A 500 kg satellite experiences a gravitational force of 3000 N, while moving
snow_lady [41]

Answer:

9.7\times 10^{-4}\ rad/s

Explanation:

Given:

m=500 kg\\F=3000 N

Radius of earth , R=6371 \times 10^3\ m\\Angular speed =\omega\\We\  know\  that\ \\F= m\times \omega^{2} \times R\\\omega^{2}=\frac{F}{m*R} \\\\=\frac{3000}{500*6371 \times 10^3\ m}

=\frac{6}{6371 \times 10^3\ m}

=9.7\times 10^{-4}\ rad/s

7 0
3 years ago
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