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FromTheMoon [43]
3 years ago
5

A battery is connected to a small torch bulb so that a current flows. The traditional direction of the flow of current and the f

low of the electrons is ... a. The same from positive to negative. b. The same from negative to positive. c. The traditional flow of current is positive to negative and the flow of electrons is the opposite. d. The traditional flow of current is negative to positive and the flow of electrons is the same.
Physics
2 answers:
EastWind [94]3 years ago
8 0

Answer:

c. The traditional flow of current is positive to negative and the flow of electrons is the opposite.

Explanation:

By convention, the electron carries negative charges, so it would flow from the negative side to the positive side. The current, by convention, would have the opposite direction, from positive to negative.

ICE Princess25 [194]3 years ago
7 0

Answer

C.The traditional flow of current is positive to negative and the flow of electrons is the opposite

Explanation:

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Andre the Giant and Tiny Tim are on ice skates in the middle of a frozen pond. Tim pushes Andre away from him
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Answer:

Absolutely the answer is ZERO

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An automobile of mass 2500 kg moving at 49.4 m/s is braked suddenly with a constant braking force of 8,868 N. How far does the c
BaLLatris [955]

Answer:

344.68 m

Explanation:

The computation of the far does the car travel before stopping is

Data provided in the question

Force = F = 8,868 N

mass = m = 2,500 kg

So,

accleration = a is

= \frac{-F}{m}\\\\\= \frac{8868}{2500}

a = -3.54 m/s^2

The initial speed = u = 49.4 m / s

final speed = v = 0

Based on the above information

Now applying the following formula

v^ 2- u^ 2= 2aS

Therefore

S = \frac{v^ 2- u^ 2}{2a}\\\\\ =  \frac{0- 49.4^ 2}{2\times -3.54}

= 344.68 m

4 0
3 years ago
Two charges, each 9 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field on the x axis at
bearhunter [10]

a) Electric field at x = -2 m: 21,060 N/C to the left

b) Electric field at x = 2 m: 18,000 N/C to the right

c) Electric field at x = 6 m: 18,000 N/C to the left

d) Electric field at x = 10 m: 21,060 N/C to the right

e) Electric field is zero at x = 4 m

Explanation:

a)

The electric field produced by a single-point charge is given by

E=k\frac{q}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

Here we have two charges of

q=9\mu C = 9\cdot 10^{-6} C

each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.

We call the charge at x = 0 as q_0 , and the charge at x = 8 m as q_8.

For a point located at x = -2 m, both the fields E_0 and E_8 produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:

E=-\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=-kq(\frac{1}{(-2)^2}+\frac{1}{(8-(-2))^2})=-21060 N/C

b)

In this case, we are analyzing a point located at

x = 2 m

The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:

E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(2)^2}-\frac{1}{(8-2)^2})=18000 N/C

And since the sign is positive, the direction is to the right.

c)

In this case, we are considering a point located at

x = 6 m

The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:

E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(6)^2}-\frac{1}{(8-6)^2})=-18000 N/C

And the negative sign indicates that the electric field in this case is towards the left.

d)

In this case, we are considering a point located at

x = 10 m

This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:

E=\frac{kq_0}{(0-x)^2}+\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(10)^2}+\frac{1}{(8-(10))^2})=21060 N/C

And the positive sign means the field is to the right.

e)

We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be

E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(-x)^2}-\frac{1}{(8-x)^2})=0

This means that we have to solve the equation

\frac{1}{x^2}-\frac{1}{(8-x)^2}=0

Re-arranging it,

\frac{1}{x^2}-\frac{1}{(8-x)^2}=0\\\frac{(8-x)^2-x^2}{x^2(8-x)^2}=0

So

(8-x)^2-x^2=0\\64+x^2-16x-x^2=0\\64-16x=0\\64=16x\\x=4 m

So, the electric field is zero at x = 4 m, exactly halfway between the two charges (which is reasonable, because the two charges have same magnitude)

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0
3 years ago
Read 2 more answers
Placing the north pole of a magnet near the south pole of another magnet results in:
MakcuM [25]

Answer:

An attractive force between the magnets. A is the correct answer if it was the north pole facing another north pole magnet the answer would have been a repulsive force between the magnets because the north pole of a magnet does not attract instead it separates from each other which is a repulsive force the best answer is A.

Explanation:

the southern pole of a magnet is a positive force while the northern pole of a magnet is negative .

Note: Magnet can attract each other only when they are facing the opposite directions a north pole will attract a south pole; the magnets pull on each other. But the two north poles will push each other away.

8 0
3 years ago
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