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Alexeev081 [22]
2 years ago
10

Which of the following changes to the earth-sun system would reduce the magnitude of the force between them to one-fourth the va

lue found in Part A?
Reduce the mass of the earth to one-fourth its normal value.
Reduce the mass of the sun to one-fourth its normal value.
Reduce the mass of the earth to one-half its normal value and the mass of the sun to one-half its normal value.
Increase the separation between the earth and the sun to four times its normal value.
Physics
1 answer:
Anna35 [415]2 years ago
4 0

Answer:

-Reduce the mass of the earth to one-fourth its normal value.

-Reduce the mass of the sun to one-fourth its normal value.

-Reduce the mass of the earth to one-half its normal value and the mass of the sun to one-half its normal value.

Explanation:

The force (F) between two massive bodies (earth-sun system) is given by the following equation:

F=\frac{g*M_{e}*M_{s}}{r^{2}}

Where "g" is the gravitational constant "Me" is Earth's mass, "Ms" is the Sun's mass and "r" is the separation between the Earth and the Sun.

1-) If we reduce the mass of the earth to one-fourth its normal value:

F^{*} =\frac{g*\frac{M_{e}}{4} *M_{s}}{r^{2}}\\F^{*} =\frac{g*{M_{e}} *M_{s}}{4r^{2}}\\F^{*} =\frac{F}{4}

2-) Reduce the mass of the sun to one-fourth its normal value.

F^{*} =\frac{g*\frac{M_{s}}{4} *M_{e}}{r^{2}}\\F^{*} =\frac{g*{M_{s}} *M_{e}}{4r^{2}}\\F^{*} =\frac{F}{4}

3-) Reduce the mass of the earth to one-half its normal value and the mass of the sun to one-half its normal value.

F^{*} =\frac{g*\frac{M_{s}}{2}*\frac{M_{e} }{2}}{r^{2}}\\F^{*} =\frac{g*{M_{s}} *M_{e}}{(2*2)r^{2}}\\F^{*} =\frac{F}{4}

4-) Increase the separation between the earth and the sun to four times its normal value.

F^{*}=\frac{g*M_{e}*M_{s}}{(4r)^{2}}\\F^{*}=\frac{g*M_{e}*M_{s}}{16r^{2}}\\F^{*}=\frac{F }{16}\\.

Therefore, changes 1, 2 and 3 would reduce the magnitude of the force between them to one-fourth of the original value.

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1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C outward

3)  y-component of the electric field at r =8.7 cm: 0

4)  x-component of the electric field at r = 1.15 cm: 1.28\cdot 10^7 N/C outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found  by using

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

E_y=0

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

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