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Dmitry [639]
3 years ago
9

Two obiect accumulated a charge of

Physics
1 answer:
tamaranim1 [39]3 years ago
3 0

Answer:

A. 181.24 N

Explanation:

The magnitude of hte electrostatic force between two charged objects is given by the equation

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

In this problem, we have:

q_1=4.5\mu C=4.5\cdot 10^{-6}C is the magnitude of the 1st charge

q_2=2.8\mu C=2.8\cdot 10^{-6}C is the magnitude of the 2nd charge

r = 2.5 cm = 0.025 m is the separation between the charges

Therefore, the magnitude of the electric force is:

F=\frac{(9\cdot 10^9)(4.5\cdot 10^{-6})(2.8\cdot 10^{-6})}{(0.025)^2}=181.44 N

So, the closest answer is

A) 181.24 N

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The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

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Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

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(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

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<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

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<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

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