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galina1969
3 months ago
11

A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri

al, a horizontal, frictionless surface is positioned next to the 12-cm-diameter sphere of a Van de Graaff generator, and a small, 5.0 g plastic cube is placed on the surface with its center 2.0 cm from the edge of the sphere. The cube is given a positive charge, and then the Van de Graaff generator is turned on, charging the sphere to a potential of 200,000 V in a negligible amount of time. How much charge does the plastic cube need to achieve a final speed of a mere 3.0 m/s? Does this seem like a practical projectile launcher?
Physics
1 answer:
Maru [3.3K]3 months ago
7 0
The charge on the plastic cube is determined as follows.
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The total duration from the stone's drop to the echo being heard is 8.9 seconds. The sound takes 0.9 seconds to reach the listener, meaning it takes the stone 8 seconds to descend to the well's bottom. Using gravitational acceleration, the well's depth calculates to be approximately 313.6 meters.
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3 months ago
A conducting sphere 45 cm in diameter carries an excess of charge, and no other charges are present. You measure the potential o
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Answer:

The excess charge is Q = 3.5 *10^{-7} \ C

Explanation:

According to the question, we are informed that

The diameter is d = 45 \ cm = 0.45 \ m

The potential of the surface is V = 14 \ kV = 14 *10^{3} \ V

The radius of the sphere is

r = \frac{d}{2}

by plugging in given values

r = \frac{0.45}{2}

r = 0.225 \ m

The potential at the surface is mathematically expressed as

V = \frac{k * Q }{r }

Where k is Coulomb's constant with a value k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

Based on the question stating there are no other charges, Q represents the excess charge

Therefore

Q = \frac{V* r}{ k}

inserting the numerical values

Q = \frac{14 *10^{3} 0.225}{ 9*10^9}

Q = 3.5 *10^{-7} \ C

7 0
2 months ago
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