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marishachu [46]
3 years ago
10

Fg =G m1m2/r2 solver for G

Physics
1 answer:
Blababa [14]3 years ago
6 0

G = \frac{F_{g} r^{2}  }{m_{1} m_{2}  }

Explanation:

 Solving for G simply implies that we make G the subject of the formula:

  Given equation:

               F_{g} = G \frac{m_{1}  m_{2} }{ r^{2} }

 To make G the subject of this expression follow these steps:

            Multiply both sides of the equation by r^{2}

   

     F_{g} x r^{2} = G \frac{m_{1}  m_{2} }{ r^{2} } x r^{2}

 

      This gives:

                F_{g}  r^{2} = G m_{1}  m_{2}

     

    Multiply both sides by \frac{1}{m_{1} m_{2} }

  F_{g}  r^{2}  x   \frac{1}{m_{1} m_{2} } =  \frac{1}{m_{1} m_{2} } x G m_{1}  m_{2}

  Therefore:

                G = \frac{F_{g} r^{2}  }{m_{1} m_{2}  }

Learn more:

Solving formula brainly.com/question/2998489

#learnwithBrainly

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m*g*h=(1/2)*m*v², masses cancel out,

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A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang
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Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: s_c = \omega_ct = 0.233t

For the horse: s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

s_c = s_h

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t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}

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3 years ago
A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th
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Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

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m=\rho\cdot V

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\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}

T = \dfrac{\rho\cdot V}{l} v^{2}

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

V = \pi \dfrac{d^{2}}{4} l

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T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2

T = 11159.4186\ldots \text{ N} = 11000 \text{ N}

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