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kicyunya [14]
3 years ago
11

What is the force felt by the electrons and the nuclei in the rod when the external field described in the problem introduction

is applied? (Ignore internal fields in the rod for the moment.)A) Both electrons and nuclei experience a force to the right.
B) The nuclei experience a force to the right and the electrons experience a force to the left.
C) The electrons experience a force to the left but the nuclei experience no force.
D) The electrons experience no force but the nuclei experience a force to the right.
Physics
1 answer:
evablogger [386]3 years ago
6 0

Answer: (B). nuclie experiences a force to the right and electron experiences a force to the left.

Explanation: The nuclie is positively charged and so it will experience a force in the same direction as the direction of the Electric Field {in this case to the right}

On the otherhand, the electrons which are negatively charge will experience a force in opposite direction to that experienced by the nuclie which is positively charged. So the electron experiences a force to the left.

I believe this is clear.

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What do we mean by the observable universe?
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3 years ago
After getting a haircut, Joey’s barber spins him around in his barber’s chair 2 times per second. Is period or frequency given?
Rzqust [24]

Answer:

Explanation:

This figure given is the frequency; 2 times per second represents frequency.

What is frequency?

  • It is the number of times per seconds something goes past or around another.

 it is expressed as:

            Frequency  = \frac{n}{t}

                  where n is the number of turns

                              t is the time taken

  Therefore, the Barber spinned him 2 times in 1 second.

The period is the inverse of frequency. It is the time taken for a body to go through a point;

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7 0
4 years ago
Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

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