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joja [24]
3 years ago
15

Why centre of mass equal to centre of gravity

Physics
2 answers:
Maksim231197 [3]3 years ago
4 0
Because mass and distance determine gravity, so the more mass you have, the more gravity.
melisa1 [442]3 years ago
4 0
Hopefully this helps you The center of gravity is based on weight, whereas the center of mass is based on mass. So, when the gravitational field across an object is uniform, the two are identical. ... More practically, the COG is the point over which the object can be perfectly balanced; the net torque due to gravityabout that point is zero.
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When did diesel locomotives replace steam locomotives?
liubo4ka [24]
To keep the energy going every day
7 0
3 years ago
An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r
bixtya [17]

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

#LearnwithBrainly

4 0
3 years ago
Explain the relationship between temperature, energy, and motion of particles in an object.
ANEK [815]

Answer:

Explanation:

Temperature is the degree of hotness or coldness of a body.

Energy is the ability to do work by a body. They are of two forms, potential and kinetic energy. Potential energy is due to the position of a body whereas kinetic energy is due to the motion of a body.

Motion is the change in position of a body with time.

Temperature, energy and motion are all related.

Thermal energy is a form of kinetic energy which is concerned about the motion particles. This form of energy results from heat changes in a body which causes temperature differences.

When a body is heat and changes temperature, the particles begins to vibrate as they gain, thermal energy, a form of kinetic energy. At a point, the particles will break lose and set in motion.

7 0
3 years ago
Can someone please help me get this right pleaseee I’ll mark brainless .
Katarina [22]

Answer:i think it is c

Explanation:

4 0
3 years ago
Read 2 more answers
A parallel-plate capacitor is connected to a battery and becomes fully charged. The capacitor is then disconnected, and the sepa
AnnZ [28]

Answer:

Increases

Explanation:

The expression for the capacitance is as follows as;

C=\frac{\epsilon _{0}A}{d}

Here, C is the capacitance, \epsilon _{0} is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.

It can be concluded from the above expression, the capacitance is inversely proportional to the distance. According to the given problem, the capacitor is disconnected from the battery and the distance between the plates is increased. Then, the capacitance of the given capacitor will decrease in this case.

The expression for the energy stored in the parallel plate capacitor is as follows;

E=\frac{Q^{2}}{2C}

Here, E is the energy stored in the capacitor, C is the capacitance and Q is the charge.

Energy stored in the given capacitor is inversely proportional to the capacitor. The charge on the capacitor is constant. In the given problem, as the distance between the parallel plates is being separated, the energy stored in this capacitor increases.

Therefore, the option (c) is correct.

8 0
3 years ago
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