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Marysya12 [62]
1 year ago
8

Romeo and Juliet are sitting on a balcony 1.5 meters apart. If Romeo has a mass of 61.6 Kg and Juliet has a mass of 48.8 kg. Wha

t is the attractive force between them?
Physics
1 answer:
Setler79 [48]1 year ago
4 0
\begin{gathered} m_{Romeo}=61.6\text{ kg} \\ m_{Juliet}=48.8\text{ kg} \\ r=1.5\text{ m} \\ G=6.673x10^{-11}Nm^2/kg^2 \\ F=G*\frac{m_{Juliet}m_{Romeo}}{r^2} \\  \\ F=\left(6.673x10^{-11}Nm^2/kg^2\right)\frac{\left(48.8\text{ kg}\right)\left(61.6\text{ kg}\right)}{\left(1.5\text{ m}\right)^2} \\  \\ F=8.91x10^{-8}N \\ The\text{ attractive force is 8.91x10}^{-8}\text{N} \end{gathered}

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Answer:

vi = 4.77 ft/s

Explanation:

Given:

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Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

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             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

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                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

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4 0
4 years ago
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Answer:

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