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dmitriy555 [2]
4 years ago
10

A ring of charge with uniform charge density is completely enclosed in a hollow donut shape. An exact copy of the ring is comple

tely enclosed in a hollow sphere.
A) What is the ratio of the flux out of the donut shape to that out of the sphere?
Physics
1 answer:
Naily [24]4 years ago
3 0

Answer:

Explanation:

According to Gauss's theorem , total flux coming out of a closed surface area is equal to 1/ε₀ times charge enclosed

This flux coming out of enclosed charge does not depend upon the shape of enclosing surface area . It only depends upon the charge enclosed .

It the present context , in both the case ie in case of donut or sphere , since amount of charge enclosed  is same , flux coming out too will be same in both the case

Hence the required ratio = 1 : 1

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A small steel roulette ball rolls around the inside of a 30 cm diameter roulette wheel. It is spun at 150 rpm, but is slows to 6
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Solution :

Given

Diameter of the roulette ball = 30 cm

The speed ball spun at the beginning = 150 rpm

The speed of the ball during a period of 5 seconds = 60 rpm

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Un cubo de madera de densidad 0.780 g/cm³ mide 11.2 cm en un lado. Cuando se coloca en agua, ¿qué altura del bloque flotará sobr
Stolb23 [73]

Answer:

2.464 cm above the water surface

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We calculate the weight of the block multiplying its density (0.78 gr/cm^3) times its volume (11.2^3  cm^3):

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Now the displaced water will have a volume equal to the base of the cube (11.2 cm^2) times the part of the cube (x) that is under water. Recall as well that the density of water is 1 gr/cm^3.

So the weight of the volume of water displaced is:

weight of water = 1 * 11.2^2 * x

we make both weight expressions equal each other for the floating requirement:

0.78 * 11.2^3 = 11.2^2 * x

then x = 0.78 * 11.2 cm = 8.736 cm

This "x" is the portion of the cube under water. Then to estimate what is left of the cube above water, we subtract it from the cube's height (11.2 cm) as follows:

11.2 cm - 8.736 cm = 2.464 cm

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