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kogti [31]
4 years ago
14

A parallel plate capacitor is connected to a battery for charging. after some time, while the battery is still connected to the

capacitor, the distance between the capacitor plates is doubled.
which of the following statements are true ?

a) the charge on the plates does not change.

b) the capacitance doubles.

c) the potential difference of the battery is halved.

d) the electric field between the plates is halved.

e) the potential difference across the plates does not change.

f) none of the above.
Physics
1 answer:
Vitek1552 [10]4 years ago
7 0
The best and most correct answer among the choices provided by your question is the fourth choice or letter D.

When the distance between the plates is doubled, <span>the electric field between the plates is halved.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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What is the effect on the force of gravity between two objects if the mass of one object remains unchanged while the distance to
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Answer:

The force of gravity after you double the mass and the distance is half of the initial force: F_{2}=\frac{1}{2}F_{1}

Explanation:

The initial force of gravity is:

F_{1}=\frac{Gm_{1}m_{2}}{r^2}

where G is the universal gravitational constant, m_{1} is the mass of the first object, m_{2} is the mass of the second object, and r is the distance between the objects.

If the mass of the second object is doubled, now we have 2m_{2}, and if the distance between the objects is also doubled instead of r now we have 2r.

So the force of gravity now is:

F_{2}=\frac{Gm_{1}(2m_{2})}{(2r)^2}\\ F_{2}=\frac{2Gm_{1}m_{2}}{4r^2} \\F_{2}=\frac{1}{2} \frac{Gm_{1}m_{2}}{r^2}

and we know that F_{1}=\frac{Gm_{1}m_{2}}{r^2}

so the new force of gravity is:

F_{2}=\frac{1}{2}F_{1}

The force of gravity after you double the mass and the distance is half of the initial force.

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3 years ago
A 25-ohm resistor at steady state draws a current of 10.8 amperes. its temperature is 310 k; the temperature of the surroundings
MA_775_DIABLO [31]

The total rate of entropy generation is 0.268 J/K.

Entropy, the degree of a gadget's thermal electricity consistent with unit temperature that is unavailable for doing useful paintings. due to the fact work is acquired from ordered molecular motion, the quantity of entropy is likewise a measure of the molecular sickness, or randomness, of a machine.

Resistance of resistor R = 25 ohm

the total rate of entropy generation ?

Current flowing through resistor I = 10 A

Heat generation due to resistance , Q = I2R = 100*25 = 2500 J

Resistor temperature T = 310 K

Surrounding temperature Ts = 300 K

Entropy change for Resistor, heat lost due to the resistance to the surrounding then Q = -ve

(ΔS)syst = - Q/T = - 2500 J/310K = - 8.064 J/K

Entropy change for the system: here heat gain from the resistor, Q = +ve

(ΔS)surr = +Q/Ts = 2500J/300K = 8.333 J/K

Total Entropy generation :

(ΔS)tot = (ΔS)syst + (ΔS) Surr = -8.064 + 8.333 = 0.268 J/K

Learn more about entropy here:- brainly.com/question/6364271

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