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barxatty [35]
4 years ago
13

Which of the following is NOT a component of Newton’s first law of motion?

Physics
1 answer:
sashaice [31]4 years ago
5 0

Answer:

C.

Explanation:

An object in stays in motion with the same speed and in the same direction unless acted upon by an <u>unbalanced</u> force.

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Which of the following are examples of acceleration?
attashe74 [19]

Answer:

cough

Explanation:

3 0
3 years ago
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A man standing in front of a mountain beats a drum at regular intervals. The rate of drumming is generally increased and he find
Likurg_2 [28]
I think it is d I hope this help you if not let me know if it is not right
3 0
4 years ago
A car is rounding a 100-m-radius curve at 25 m/s.What is the minimum possible coefficient of static friction between the tires a
Crazy boy [7]

Answer:

The minimum possible coefficient of static friction between the tires and the ground is 0.64.

Explanation:

if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :

Fc = f

m×(v^2)/(R) = μ×m×g

    (v^2)/(R) = g×μ

               μ = (v^2)/(R×g)

                  =  ((25)^2)/((100)×(9.8))

                  = 0.64

Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.

4 0
3 years ago
Please help!
Trava [24]

Answer:

3 N to the right

Explanation:

There are two forces acting on the car:

- A force of 10 N towards the right

- A force of 7 N towards the left

Therefore, the net force is given by the difference between the two, since they are in opposite directions:

F=10 N-7 N=3 N

And the direction is to the right, since the force to the right has greater magnitude than the force to the left.

7 0
3 years ago
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A race car moving along a circular track has a centripetal acceleration of 15.4 m/s? If the car has
Helen [10]

Answer:

r = 58.44 [m]

Explanation:

To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.

a = v²/r

where:

a = centripetal acceleration = 15.4 [m/s²]

v = tangential speed = 30 [m/s]

r = radius or distance [m]

r = v²/a

r = 30²/15.4

r = 58.44 [m]

3 0
3 years ago
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