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svet-max [94.6K]
3 years ago
14

Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the bat

tery connected).
a. Capacitance
b. Charge on the plates
c. Voltage across the plates
d. Net electric field between the plates
e. Energy stored in the capacitor
Physics
1 answer:
RSB [31]3 years ago
3 0

Answer:

a. Capacitance

b. Charge on the plates  

e. Energy stored in the capacitor

Explanation:

Let A be the area of the capacitor plate

The capacitance of a capacitor is given as;

C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\

where;

V is the potential difference between the plates

The charge on the plates is given as;

Q = \frac{V\epsilon _0 A}{d}

The energy stored in the capacitor is given as;

E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2

Thus, the physical variables listed that will change include;

a. Capacitance

b. Charge on the plates  

e. Energy stored in the capacitor

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An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

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Answer:

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 m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear m_{ear} = 0.030 \ kg

Farther distance between one ear and the black hole (d) = 6.0 cm

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Closer distance between the other ear and the black home is (d) 6.0 cm

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NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R -  d) \omega^2

\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)

The net force on the ear farther to the black hole is :

\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R +  d) \omega^2

\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

T = \frac{3GMm_{ear}d}{R^3}

T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}

T = 2073.9 N\\T = 2.07 KN

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The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

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Q = 39.46 ft³/hr

3 0
3 years ago
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