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lubasha [3.4K]
3 years ago
14

A small object has charge Q. Charge q is removed from it and placed on a second small object The two objects are placed I m apar

t. For the force that each object exerts on the other to be a maximum, what should q be? a) q 20 c) q Q/2 d) q /4
Physics
1 answer:
Whitepunk [10]3 years ago
4 0

Answer:

q = Q/2

option c is correct

Explanation:

given data

charge = Q

distance r = 1 m

charge = q

to find out

what should q be

solution

we have given charge q is remove so

q1 = Q-q

q2 = q

we know by coulombs law force

force = kq1q2 / r²   .............1

put here value and we know k is constant

F = k (Q-q) q / 1

now take derivative charge q and put = 0

dF/dq = k d(Q-q)q / dq

so  k (Q-q)q =0

Q = 2q

q = Q/2

so option c is correct

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Answer:

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Explanation:

To find the required force you first calculate the angle of the ramp, by using the following relation:

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Next, you use the Newton second law to know what is the x component (in a rotated coordinate system) of the gravitational force:

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the required force must be, at least, the last force Fx. You know that the weight of the object is 490N = mg. Hence, you have:

F=F_x=(450N)sin(23.57\°)=180N

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- - - - - - - - - - - - - - - - - - - - - - - -

TRANSLATION:

Para encontrar la fuerza requerida, primero calcule el ángulo de la rampa, utilizando la siguiente relación:

Luego, usa la segunda ley de Newton para saber cuál es el componente x (en un sistema de coordenadas girado) de la fuerza gravitacional:

la fuerza requerida debe ser, al menos, la última fuerza Fx. Sabes que el peso del objeto es 490N = mg. Por lo tanto, tienes:

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