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Drupady [299]
3 years ago
9

A 70 pF capacitor and a 280 pF capacitor are both charged to 3.6 kV. They are then disconnected from the voltage source and are

connected together, positive plate to positive plate and negative plate to negative plate.
Find the resulting potential difference across each capacitor.
kV (70 pF capacitor)
kV (280 pF capacitor)
Find the energy lost when the connections are made.J
Physics
1 answer:
e-lub [12.9K]3 years ago
6 0

Answer:

0.636 kJ

Explanation:

The charge on any capacity, q = CV, thus,

The initial charge on the 70 pF capacitor is

q = Cv

q = 70*10^-12 * 3.6*10^3

q = 2.52*10^-7 C

The charge on the 280 pF capacitor is q = C*v

q = 280*10^-12 * 3.6*10^3

q = 1.008x10^-6 C

When they are connected as stated, the net total charge remaining will be 1.008*10^-6 - 2.52*10^-7 = 7.56*10^-7 C

Since the capacitors are in parallel, the equivalent capacitance will be 70 + 280 pF = 350 pF

Remember, q = CV, then V = q/C

V = 7.56*10^-7 C / 350*10^12 F

V = 2160 V

b) The energy before is 1/2 C*v²

E = 1/2 * 70*10^-12 * 3600² + 1/2 * 280*10^-12 * 3600²

E = 4.536*10^-4 J + 1.814*10^-3 J

E = 2.268 kJ

The energy After is 1/2 Cv²

E = 1/2 * 70*10^-12 * 2160² + 1/2 * 280*10^-12 * 2160²

E = 3.266*10^-4 J + 1.306*10^-3 J

E = 1.632 kJ

so the loss is 2.268 - 1.632 = 0.636 kJ

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A block of gelatin is 120mm by 120mm by 40mm whrn unstressed. A force of 49N is applied tangentially to the upper surface causin
Inessa05 [86]

Answer:

The shearing stress is 10208.3333 Pa

The shearing strain is 0.25

The shear modulus is 40833.3332 Pa

Explanation:

Given:

Block of gelatin of 120 mm x 120 mm by 40 mm

F = force = 49 N

Displacement = 10 mm

Questions: Find the shear modulus, Sm = ?, shearing stress, Ss = ?, shearing strain​, SS = ?

The shearing stress is defined as the force applied to the block over the projected area, first, it is necessary to calculate the area:

A = 40*120 = 4800 mm² = 0.0048 m²

The shearing stress:

Ss=\frac{F}{A} =\frac{49}{0.0048} =10208.3333Pa

The shearing strain is defined as the tangent of the displacement that the block over its length:

SS=tan\theta =\frac{Displacement}{L}  =\frac{10}{40} =0.25

Finally, the shear modulus is the division of the shearing stress over the shearing strain:

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tino4ka555 [31]

Angle, θ2 at which the light leaves mirror 2 is 56°

<u>Explanation:</u>

Given-

θ1 = 64°

So, α will also be 64°

According to the figure:

α + β = 90°

So,

β = 90° - α

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γ = 180° - 120° - β

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γ = 34°

γ + δ = 90°

δ = 90° - γ

δ = 90° - 34°

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According to the law of reflection,

angle of incidence = angle of reflection

θ2 = δ = 56°

Therefore, angle θ2 at which the light leaves mirror 2 is 56°

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