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katovenus [111]
3 years ago
6

Suppose a piece of dust finds itself on a CD. If the spin rate of the CD is 500 rpm, and the piece of dust is 4.3 cm from the ce

nter, what is the total distance traveled by the dust in 3 minutes? (Ignore accelerations due to getting the CD rotating.)
Physics
1 answer:
labwork [276]3 years ago
8 0

Answer:

405.3m

Explanation:

Since the CD spin rate is 500 rpm, or revolution per minutes, its angular speed in rad per second is

\omega = 500 rev/min * 2\pi rad/rev * 1/60 min/sec = 52.36 rad/s

The dusk is 4.3 cm from center, so its velocity must be

v = R\omega = 4.3 * 52.36 = 225.14 cm/s

Then the distance traveled by the dusk after 3 minutes, or 180 seconds is

d = v*t = 225.14 * 180 = 40526.54 cm or 405.3m

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A spherical capacitor contains a charge of 3.40 nC when connected to a potential difference of 240.0 V. Its plates are separated
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Answer:

A) 1.4167 × 10^(-11) F

B) r_a = 0.031 m

C) E = 3.181 × 10⁴ N/C

Explanation:

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Potential difference;V = 240 V

Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m

A) The formula for capacitance is given by;

C = Q/V

C = (3.4 × 10^(-9))/240

C = 1.4167 × 10^(-11) F

B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.

C = (4πε_o)/(1/r_a - 1/r_b)

Rearranging, we have;

(1/r_a - 1/r_b) = (4πε_o)/C

ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m

Plugging in the relevant values, we have;

(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))

(1/r_a) - 24.3902 = 7.8501

1/r_a = 7.8501 + 24.3902

1/r_a = 32.2403

r_a = 1/32.2403

r_a = 0.031 m

C) Formula for Electric field just outside the surface of the inner sphere is given by;

E = kQ/r_a²

Where k is a constant value of 8.99 × 10^(9) Nm²/C²

Thus;

E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²

E = 3.181 × 10⁴ N/C

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1 year ago
What is the relationship between current and voltage in the filament of an incandescent light bulb?
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