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3 years ago

What does the number, or density, of field lines on a source charge indicate?

Physics

2 answers:

Harrizon [31]3 years ago

6 0

Hello.

The answer: C. amount of charge on the source charge

The field lines all represent the movements of field charges. The more crowded the field lines or the more density of electric field the higher is the magnitude of the source charge.

Have a nice day.

Natali [406]3 years ago

3 0

The number or density of field lines on a source charge indicates the amount of charge on the source charge.

Answer: Option C

<u>Explanation: </u>

The source charge is the point from which the charge lines or the lines of energy will be started. So the number of lines emitted from the point charge or the source charge will indicate the strength of the source charge.

Also by measuring the density of the source charge i.e., the number of lines emitted per unit volume or the field lines which is the measure of lines emitted per unit area also indicate the strength of the source charge. So these strength are nothing but the amount of charge present in the source charge.

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13. Austin rode his bike 10 m/s for two minutes. How far did he travel? A. 200 meters B. 1200 meters C. 1000 meters D. 20 meters

Semmy [17]

Answer:

B. 1200

Explanation:

60 sec in one min in 2 min there will be 120 sec. 10x120=1200

5 0

3 years ago

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

5 0

3 years ago

How much work is needed to lift a 3kg create a vertical displacement of 22m

Ulleksa [173]

I think its W= 647.239J

6 0

3 years ago

Can someone help me?!!!!!

ladessa [460]

Answer:

magnitude: 21.6; direction: 33.7 degrees

Explanation:

When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have

vector: (-3,-2)

Scalar: -6

so the vector multiplied by the scalar will have components

$(-3\cdot (-6), -2 \cdot (-6))=(18,12)$

The magnitude is given by Pythagorean's theorem:

$m=\sqrt{18^2+12^2}=21.6$

and the direction is given by the arctan of the ratio between the y-component and the x-component:

$\theta = tan^{-1} (\frac{12}{18})=33.7^{\circ}$

3 0

3 years ago

RANGKAIAN RLC SERI PADA TEGANGAN BOLAK BALIK 220V R=80 OHM XL= 90 OHM XC = 30 OHM MAKA TEGANGAN PADDA INDUKTOR ADALAH???

Whitepunk [10]

Okay, 90% of this is nonsense besides the numbers maybe.

8 0

3 years ago