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nydimaria [60]
2 years ago
13

A 6.0 C charge is placed at the origin and a second charge is placed on the x-axis at x 0.3m. If the resulting electric force on

the second charge is 5.4 N in the positive x-direction, what is the value of its charge?
Physics
1 answer:
almond37 [142]2 years ago
8 0

Answer:

9 x 10^-12 c

Explanation:

q1 = 6 C

d = 0.3 m

q2 = ?

F = 5.4 N

As te force on the second charge is in the positive X axis direction so the charge on the second particle is positive in nature.

F = K q1 q2 / d^2

5.4 = ( 9 x 10^9 x 6 x q2) / (0.3 x 0.3)

5.4 x 0.3 x 0.3 = 54 x 10^9 x q2

q2 = 9 x 10^-12 C

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When have passed 3.9[s], since James threw the ball.

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y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\

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Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

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