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PilotLPTM [1.2K]
3 years ago
12

1 oz = _____________ mg

Physics
1 answer:
Luba_88 [7]3 years ago
4 0

Answer:

28349.5 Mg

Explanation:

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Gauss’ law: a. Relates the surface charge density to the electric field.b. Relates the electric field at points on a closed surf
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Answer:

b. Relates the electric field at points on a closed surface to the net charge enclosed by that surface

Explanation:

Gauss's law states that the flux of certain fields through a closed surface is proportional to the magnitude of the sources of that field within the same surface. The electric flux expresses the measure of the electric field that crosses a certain surface. Therefore, the electric field on a closed surface is proportional to the net charge enclosed by that surface.

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Dark matter is missing stars that have fallen from the sky.<br> true or false
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The answer to this question is False

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A student performs an exothermic reaction in a beaker and measures the temperature. If the thermometer initially reads 35 degree
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B 25 degrees Celsius

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The resistivity of gold is at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 m
IgorC [24]

Answer:

0.0360531138247 V/m

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d = Diameter = 0.9 mm

A = Area = \dfrac{\pi}{4}d^2

E = Electric field

Resistivity is given by

\rho=\dfrac{EA}{I}\\\Rightarrow E=\dfrac{\rho I}{A}\\\Rightarrow E=\dfrac{2.44\times 10^{-8}\times 940\times 10^{-3}}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}\\\Rightarrow E=0.0360531138247\ V/m

The  electric field in the wire is 0.0360531138247 V/m

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4 years ago
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14. The design for a rotating spacecraft below consists of two rings. The outer ring with a radius of 30 m holds the living quar
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Answer:

T= 11.0003s

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From the question we are told that

The outer ring with a radius of 30 m

inner Gravity Approximately 9.80 m/s'

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Centripetal acceleration enables Rotation therefore?

     \omega ^2 r =Angular\ acc

Considering the outer ring,

 \omega ^2 r = 9.8

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 \omega = \sqrt{\frac{9.8}{30}}

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Therefore solving for  Period T

Generally the equation for solving Period T is mathematically given as

 T= \frac{2\pi}{\omega}

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