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kolezko [41]
3 years ago
11

State the Newton 2nd law of motion and also prove that F= ma​

Physics
1 answer:
saw5 [17]3 years ago
3 0

Explanation:

F = ma is the formula of Newton's Second Law of Motion. Newton's Second Law of Motion is defined as Force is equal to the rate of change of momentum. For a constant mass, force equals mass times acceleration.

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What type of energy slows down
Tema [17]
It is Potential energy it's at rest
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3 years ago
What force does it take to accelerate a 7.2 kg object 3.0 m/s^2.
grin007 [14]

Answer:

<h2>21.6 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 7.2 × 3 = 21.6

We have the final answer as

<h3>21.6 N</h3>

Hope this helps you

3 0
2 years ago
An object is moving with the speed of light around the Earth. How much time will it take to complete one round trip along the eq
dimulka [17.4K]

Answer:

0.14 seconds

Explanation:

The speed of light in vacuum is approximately 3.0*10^8. The distance that would be covered by the object would be equivalent to the circumference of the cross-section of the earth on the equator.

Circumference = 2\pi*6400000 =4.02*10^7

Time = distance/speed = 4.2*10^7 / 3.0*10^8 =0.14s

8 0
3 years ago
A brown bear runs at a speed of 9.0\,\dfrac{\text m}{\text s}9.0 s m ​ 9, point, 0, start fraction, start text, m, end text, div
KengaRu [80]

Explanation:

Below is an attachment containing the solution.

7 0
3 years ago
Read 2 more answers
A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axles. A light cord wrapped around
Alex_Xolod [135]

Answer:

\alpha =\frac{m*g*R}{I-m*R^2}

a = \frac{m*g*R^2}{I-m*R^2}

T=\frac{I*m*g}{I-m*R^2}

Explanation:

By analyzing the torque on the wheel we get:

T*R=I*\alpha    Solving for T:   T=I/R*\alpha

On the object:

T-m*g = -m*a    Replacing our previous value for T:

I/R*\alpha-m*g = -m*a

The relation between angular and linear acceleration is:

a=\alpha*R

So,

I/R*\alpha-m*g = -m*\alpha*R

Solving for α:

\alpha =\frac{R*m*g}{I+m*R^2}

The linear acceleration will be:

a =\frac{R^2*m*g}{I+m*R^2}

And finally, the tension will be:

T =\frac{I*m*g}{I+m*R^2}

These are the values of all the variables: α, a, T

8 0
3 years ago
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