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Alex17521 [72]
3 years ago
15

When a certain capacitor carries charge of magnitude Q on each of its plates, it stores energy Ep. In order to store twice as mu

ch energy, how much charge should it have on its plates
Physics
1 answer:
padilas [110]3 years ago
3 0
<h2>Answer:</h2>

2Q

<h2>Explanation:</h2>

When a capacitor carries some certain charge, the energy stored in the capacitor is its electric potential energy E. The magnitude of this potential energy is given by;

E  = \frac{1}{2}qV            ------------(i)

Where;

q = charge between the plates of the capacitor

V = potential difference between the plates of the capacitor

From the question;

q = Q

E = Ep

Therefore, equation (i) becomes;

Ep = \frac{1}{2} QV              ----------------(ii)

Make V subject of the formula in equation (ii)

V = \frac{2E_{p}}{Q}

Now, when the energy is doubled i.e E = 2Ep, equation (i) becomes;

2Ep = \frac{1}{2}qV

Substitute the value of V into the equation above;

2Ep = \frac{1}{2}(q *\frac{2E_{p}}{Q})

Solve for q;

2E_{p} = \frac{2qE_p}{2Q}

2E_{p} = \frac{qE_p}{Q}

q = 2Q

Therefore, the charge, when the energy stored is twice the originally stored energy, is twice the original charge. i.e 2Q

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4 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

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Answer:

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adoni [48]

Answer:

<em>The cantaloupe has a speed of 117.6 m/s</em>

Explanation:

<u>Free Fall Motion</u>

It occurs when an object falls under the sole influence of gravity. Any object that is being acted upon solely by the force of gravity is said to be in a state of free fall. Free-falling objects do not face air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is g = 9.8 m/s^2.

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The cantaloupe has been dropped from rest. We are required to find the speed after t=12 seconds.

Calculate the final speed:

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The cantaloupe has a speed of 117.6 m/s

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Nearly 14 billion years, according to astronomers. Happy to help!
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