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4 years ago

8

An isolated conducting sphere has a 12 cm radius. One wire carries a current of 1.0000049 A into it while another wire carries a

current of 1.0000000 A out of it. How long in seconds would it take for the sphere to increase in potential by 1500 V

1 answer:

tatuchka [14]4 years ago

7 0

Answer:

it will take for the sphere to increase in potential by 1500 V, 503.71 s.

Explanation:

The charge on the sphere after t seconds is:

q = (1.0000049 - 1.0000000) t = 0.0000049 t

The voltage on the surface is

V = k * $\frac{q}{R}$ = k 0.0000049 t / R

solve for t

t = (R*V) / (0.0000049 k) = (0.12 * 1500) / (0.0000049 * $9e^{9}$ ) = 503.71 s

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A tennis racket strikes a tennis ball with a force of 100 N. With what force

viva [34]

Answer:

-100N

Explanation:

Newton's third law of motion states that to every force exerted on one body, there is an equal and opposite force. This means that if object A exerts an ACTION force on B, there is a force called REACTION FORCE, which is equal and opposite, exerted on A by B.

The action and reaction forces are equal in size/magnitude but opposite in direction. In this case where a tennis racket strikes a tennis ball with a force (action force) of 100N, the ball will strike the racket with a reaction force of -100N.

F(RB) = -F(BR)

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3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

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As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

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Therefore the velocity when they are about to collide is $1.83*10^7m/s$

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3 years ago

A 15-kg block at rest on a horizontal frictionless surface is attached to a very light ideal spring of force constant 450 N/m. T

den301095 [7]

Answer:

0.266 m

Explanation:

Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,

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$m_pv_p=v_c(m_p+m_b)$ where sunscripts p and b represent putty and block respectively, c is common velocity.

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The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy

$0.5(m_p+m_b)v_c^{2}=0.5kx^{2}$ where k is spring constant and x is the compression of spring. Substituting the given values then

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3 years ago

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Vesnalui [34]

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R3 <<<< ===== answer.

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3 years ago

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Vinvika [58]

Answer:

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3 years ago