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otez555 [7]
4 years ago
14

The parallel plates in a capacitor, with a plate area of 8.50 cm2 and an air-filled separation of 3.00 mm, are charged by a 6.00

V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 8.00 mm. Neglect fringing. (a) Find the potential difference between the plates. (b) Find the initial stored energy.
Physics
1 answer:
ss7ja [257]4 years ago
3 0

Answer:

(A) Potential difference between plates will be 6 volt

(b) Initial energy stored will be 45.126\times 10^{-7}J

Explanation:

It is given area of parallel late capacitor A=8.50cm^2=8.5\times 10^{-4}m^2

Initial voltage V_0=6volt

Distance between plates d = 3 mm = 0.003 m

Capacitance C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-7}\times 8.5\times 10^{-4}}{0.003}=2.507\times 10^{-7}F

So charge on the capacitor q_0=CV_0=2.507\times 10^{-7}\times 6=15.042\times 10^{-7}C

(a) This charge will be constant after disconnection.

After disconnection voltage will be

V_1=\frac{q_0}{C}=\frac{15.042\times 10^{-7}}{2.507\times 10^{-7}}=6volt

(b) Initial energy stored in the capacitor

U=\frac{1}{2}CV^2=\frac{1}{2}\times 2.507\times 10^{-7}\times 6^2=45.126\times 10^{-7}J

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3 years ago
When a stress is applied to rocks within the lithospere, the rocks will tend to deform ...?
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6 0
3 years ago
A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline
balandron [24]

Answer:

<em>2.78m/s²</em>

Explanation:

Complete question:

<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>

According to Newton's second law of motion:

\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\

Where:

\mu is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

gsin\theta - \mu g cos\theta = a_x\\

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>

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3 years ago
7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ba
Dima020 [189]

\\  \tt \longmapsto \: h =  - 4.8 {t}^{2}  + 8.2t + 1.8 \\ \\  \tt \longmapsto \: h =  - 4.8(1.7) {}^{2}  + 8.2 \times 1.7 + 1.8 \\ \\  \tt \longmapsto \:  - 4.8 \times 2.89 + 1.39 + 1.8 \\ \\  \tt \longmapsto \: 13.8 + 1.39 + 1.8 \\ \\  \tt \longmapsto \: 17.06

6 0
3 years ago
A sample of oxygen gas occupies a volume of 5.0L at 90kPa pressure. What volume will it occupy at 145kPa?
Burka [1]

Answer:

<h3>The answer is option A</h3>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

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We have the final answer as

<h3>3.10 L</h3>

Hope this helps you

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