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dlinn [17]
1 year ago
8

What constant acceleration is required to increase the speed of a car from 27 mi/h to 53 mi/h in 3s?

Physics
1 answer:
mario62 [17]1 year ago
7 0

Answer:

8.87 m/s^2  

Explanation:

Acceleration is change in velocity over a change in time

27 to 53 = 26 m/s   change in velocity

26 m/s / 3 s = 8.87 m /s^2

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186282 miles. Hope it helps

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Consider a point 0.5 m above the midpoint of the two charges. As you can verify by removing one of the positive charges, the ele
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Answer:

the magnitude of the total electric field is 25 V/m

Explanation:

Given data

point above = 0.5 m

charges = 18 V/m

to find out

the magnitude of the total electric field

solution

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How many hydrogen and carbon atoms in a diamond
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Answer:

Explanation:

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4 0
3 years ago
A 20 kg crate initially at rest on a horizontal floor requires a 80 N horizontal force to set it in motion. Find the coefficient
e-lub [12.9K]

Answer:

<em>The coefficient of static friction between the crate and the floor is 0.41</em>

Explanation:

<u>Friction Force</u>

When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

Fr=\mu N          [1]

Where \mu is the coefficient of static or kinetics friction and N is the normal force.

If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:

N = W = m.g

The crate of m=20 Kg has a weight of:

W = 20*9.8

W = 196 N

The normal force is also N=196 N

We can find the coefficient of static friction by solving [1] for \mu:

\displaystyle \mu=\frac{Fr}{N}

The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:

\displaystyle \mu=\frac{80}{196}

\mu=0.41

The coefficient of static friction between the crate and the floor is 0.41

7 0
3 years ago
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