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Marysya12 [62]
11 months 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]11 months 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: Option d.

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The force between charges can be expressed as:

F = k*q1*q2/r^2

where k is a constant, q1 and q2 are the charges and r is the distance between them.

We hare in a equilateral triangle, so al the distances are equal.

and the charges are:

qa = q

qb = q

qc = 2q

Now, for example, the force that experiments charge A is (in the y axis)

F = (k*qa*qb/r^2 + k*qa*qc/r^2)*cos(30°) = cos(30°)*(q + 2q)*k*q/r^2 = cos(30°)*3*k*q^2/r^2

for charge B, the force is again, in the y axis.

F = (k*qb*qa/r^2 + k*qb*qc/r^2)*cos(30°) = cos(30°)*(q + 2q)*k*q/r^2 = cos(30°)*3*k*q^2/r^2

for particle C, we have:

F = (k*qc*qa/r^2 + k*qc*qb/r^2)*cos(30°) = cos(30°)*(q + q)*k*2q/r^2 = cos(30°)*4*k*q^2/r^2

Wher the cosine of 30° comes because we have a equilatiral triangle, where all the internal angles are 60°, so if we draw a line that cuts the angle by half (our y-axis) the angles to each side are 30°.

We can do a similar process for the forces in the x-axis, and we will reach the same conclusion:

Now, this means that the force that experiences the charge C is the biggest force, so the correct option is c.

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