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3 years ago

6

A 4.55 nF parallel-plate capacitor contains 27.5 μJ of stored energy. By how many volts would you have to increase this potentia

l difference in order for the capacitor to store 55.0 μJ of potential energy?
Express your answer in volts as an integer.

1 answer:

aivan3 [116]3 years ago

5 0

Answer:

$\Delta V=V_{2}-V_{1}=45.4V$

Explanation:

The energy, E, from a capacitor, with capacitance, C, and voltage V is:

$E=\frac{1}{2} CV^{2}$

$V=\sqrt{2E/C}$

If we increase the Voltage, the Energy increase also:

$V_{1}=\sqrt{2E_{1}/C}$

$V_{2}=\sqrt{2E_{2}/C}$

The voltage difference:

$V_{2}-V_{1}=\sqrt{2E_{2}/C}-\sqrt{2E_{1}/C}$

$V_{2}-V_{1}=\sqrt{2*55*10^{-6}/4.55*10^{-9}}-\sqrt{2*27.5*10^{-6}/4.55*10{-9}}=45.4V$

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A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi

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Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn = normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

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Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

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