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Fudgin [204]
3 years ago
6

A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance,

the Coulomb force between the two is to be maximized?
Physics
1 answer:
Lisa [10]3 years ago
6 0

Answer:

Explanation:

The two charges are q and Q - q. Let the distance between them is r

Use the formula for coulomb's law for the force between the two charges

F = \frac{Kq_{1}q_{2}}{r^{2}}

So, the force between the charges q and Q - q is given by

F = \frac{K\left ( Q-q \right )q}}{r^{2}}

For maxima and minima, differentiate the force with respect to q.

\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )

For maxima and minima, the value of dF/dq = 0

So, we get

q = Q /2

Now \frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}

the double derivate is negative, so the force is maxima when q = Q / 2 .

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A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground.
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Answer:

a) \Delta U_g=-5.3kJ

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the gravitational potential energy is given by:

U_g=m.g.h\\

\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ

The kinetic energy is given by:

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the initial kinetic energy is zero because the motion started from rest, so:

K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ

applying the conservation of energy theorem:

U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ

The work done by the friction force is given by:

W_f=F_f.h.cos(\theta)\\

the angle of the force is 180 degrees because it's against the movement:

F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN

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