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hjlf
3 years ago
5

Two students suggest different functions for the battery plays a flashlight.

Physics
2 answers:
KiRa [710]3 years ago
7 0

Answer:

Student B is correct

Explanation:

The battery acts like a pump to drive the electric charge. The electric charge comes from the conducting material but the battery provides a potential difference or a gradient of flow for this charge which is necessary for the movement of the charge along the conductor.

irga5000 [103]3 years ago
3 0

Answer:

Student b

Explanation:

Because, Battery pumps the charges around the circuit. Here, charges are the free electrons. And this charges are already present in the conductor. When we connect this conductor with the battery - It set up an electric field and The diffrence between the potential between the ends of the wire pushes this free charges to move in perticular direction. (This is called electric energy)

This electric energy is converted by chemical reactions in the battery. When the battery is dead - It means that the chemical energy producing chemicals has been consumed. thereafter there will be no more chemical reactions takes place to produce electric energy. Hence we say that the battery is dead.

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On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f
Sati [7]
On the moon, the gravitational acceleration is 1/6 of 9.8 m/s², so
g = 9.8/6 = 1.633 m/s²

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Part A
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8 0
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A parallel plate capacitor is made of two large plates ofarea
Murrr4er [49]

Answer:

C = \frac{\epsilon_0 \epsilon_1 A}{\epsilon_0 t + \epsilon_1(a-t)}\\

Explanation:

If there is a dielectric slab with thickness less than the distance between the plates is inserted inside a capacitor, then that capacitor can be regarded as two capacitors connected in series.

Let's assume that the slab is placed onto the lower side. So, the capacitance of that part of the capacitor, C1 is

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The other part of the capacitor is

C_2 = \epsilon_0 \frac{A}{a-t}

Two capacitors are connected in series:

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If we know the charge of the plates, we could relate the potential V0 to capacitance via

C = \frac{Q}{V_0}

4 0
3 years ago
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