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Eduardwww [97]
3 years ago
10

A metal spherical shell with inner radius 14 cm and outer radius 24 cm has a net charge of 1= -3 nC. At the center of the shell

is a small particle with charge Q2= -7 nC. What is the charge density in (C/m2) on the outer surface of the spherical shell?
Physics
1 answer:
CaHeK987 [17]3 years ago
6 0

Answer:

I don't no the answer sorry

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If the velocity of gas molecules is doubled the its kinetic energy will be
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Answer: Quadrupled

Explanation:

Because when it’s doubled they are going to make another pair so 2+2=4

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Answer the question below
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Calculate the total mass of these letters in grams (mxe=131.29amu)
Lady_Fox [76]
We know avogadro's number = 6.023 x 10²³
1 amu = 1 atomic mass unit = 1/ (6.023 x 10²³) gm = 1.66 x 10⁻²⁴grams
<span>Now, to convert amu to grams,
 131.29 amu = 131.29 x 1.66 x 10</span>⁻²⁴grams
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4 0
4 years ago
Read 2 more answers
What are the basic si units for the wavelength of light?
noname [10]

Answer:

Meter (m)

Explanation:

The wavelenght of a light wave is a measure of the distance between two successive crests (or two successive troughs) of a light wave.

Since the SI units for the distance is the meter (m), then the SI unit for the wavelength is also the meter (m).

Wavelength is related to the frequency of the light wave by:

\lambda=\frac{c}{f}

where

c is the speed of light

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7 0
3 years ago
A neutron star is an extremely dense, rapidly spinning object that results from the collapse of a massive star at the end ofits
Westkost [7]

Answer:

(a). The rotational inertia is 5.72\times10^{39}\ kg m^2

(b). The magnitude of the magnetic torque is 3.20\times10^{35}\ N-m

Explanation:

Given that,

Mass of neutron M_{n}= 13M_{s}

Density of neutron \rho=4.8\times10^{17}\ kg/m^3

(a). We need to calculate the rotational inertia

Using formula of rotational inertia  for sphere

I=\dfrac{2}{5}MR^2...(I)

We know that,

\rho=\dfrac{M}{V}

Put the value of volume

\rho=\dfrac{3M_{n}}{4\pi R^3}

R^2=(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}

Put the value of R in equation (I)

I=\dfrac{2}{5}\times M_{n}\times(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}

Put the value into the formula

I=\dfrac{2}{5}\times(13\times2\times10^{30})^{\frac{5}{3}}\times(\dfrac{3}{4\pi\times(4.8\times10^{17})})^{\frac{2}{3}}

I=5.72\times10^{39}\ kg m^2

The rotational inertia is 5.72\times10^{39}\ kg m^2.

(b). We need to calculate the magnitude of the magnetic torque

Using formula of torque

\tau=I\times \alpha

Put the value into the formula

\tau=5.72\times10^{39}\times5.6\times10^{-5}

\tau=3.20\times10^{35}\ N-m

The magnitude of the magnetic torque is 3.20\times10^{35}\ N-m

Hence, (a). The rotational inertia is 5.72\times10^{39}\ kg m^2

(b). The magnitude of the magnetic torque is 3.20\times10^{35}\ N-m

4 0
3 years ago
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