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Mashutka [201]
3 years ago
10

In the equation for the gravitational force between two objects, what does G represent?

Physics
2 answers:
user100 [1]3 years ago
5 0

The G is represented as “Universal Gravitational constant” in the gravitational force equation acting between the two objects.

<u>Explanation: </u>

The force of gravitation is given by following equation, \bold{F= \frac{- G (M m)}{r^2}}

<em>F - Force of gravitation, </em>

<em>G - Universal gravitational constant, </em>

<em>M - Mass of 1^{st} object, </em>

<em>m - Mass of 2^{nd} object, </em>

<em>R - Distance between two objects. </em>

The negative sign here denotes that gravitational force is experienced along the line of separation of the two objects. And it also indicates that it is an attractive force between the two.

Monica [59]3 years ago
3 0
G is actually the universal gravitational constant.
6.674×10−11 m3⋅kg−1⋅s−2
This proportional to the product of the two objects masses and the inverse square of their distance.


This different from g which the local gravitational constant.
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PART A )

a_c = \frac{V^2}{R}

a_c = \frac{V^2}{350}

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 5.25ft/s^2

a = \sqrt{a_t^2+a_r^2}

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27.5625 = 1.21 + \frac{v^4}{122500}

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Now calculate the angular velocity of the motorcycle

v = r\omega

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Calculate the angular acceleration of the motorcycle

a_t = r\alpha

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Calculate the time needed by the motorcycle to reach an acceleration of

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PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

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PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a_r = \frac{v^2}{R}

a_r = \frac{21.5^2}{350}

a_r =1.3207ft/s^2

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a = \sqrt{a_t^2+a_r^2}

a = \sqrt{(1.1)^2+(1.3207)^2}

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PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

3 0
3 years ago
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