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Rom4ik [11]
3 years ago
13

Consider an electron traveling horizontally in the positive direction, above, near, and parallel to a current-carrying wire. The

current is traveling in the negative direction.
(a) Will the electron keep on going straight, move towards the wire or away from the wire?


(b) If we replace the electron with a proton and now the proton is coming from above, traveling in the perpendicular direction, moving towards the wire. Will the proton hit the wire, veer to the left? veer to the right? stop? go out of the page? go onto the page?
Physics
1 answer:
allsm [11]3 years ago
6 0

Answer:

(a) The electron will move towards the wire.

The direction of the magnetic fields created by the wire can be found via right-hand rule. If you point your thumb towards the direction of the current, and if you curl your fingers, the direction of your four fingers will give the direction of the magnetic field. In this case, magnetic field is around the wire, and into the page just above the wire, where the electron is located.

\vec{F} = q\vec{v} \times \vec{B}

According to the above formula, the direction of the force the wire applies to the electron can be found by right-hand rule.

Since the electron has a negative charge, the direction of the force is towards the wire.

(b) The proton will veer to the right.

The direction of the magnetic field is the same as the previous part. The proton has a positive charge, and coming from above. The direction of its velocity is downwards. The magnetic field above the wire is pointed into the page. Using the right-hand rule, the magnetic force on the proton is directed to the right, with respect to us.

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Bridget is riding her bicycle up a hill. Which statements are correct? Check all that apply.
Svetlanka [38]

Answer:

Bridget is transferring energy to the bicycle.

The bicycle is using energy to do work.

Bridget has kinetic energy.

The bicycle has potential energy.

The bicycle has mechanical energy.

Explanation:

Energy can be transformed from one form to another. A body possess kinetic energy due to virtue of its motion. Potential energy is possessed by a body due to virtue of its position. mechanical energy is the sum of potential energy and kinetic energy. Nuclear energy is produced when atoms split or two atoms fuse together.

When Bridget is riding bicycle up a hill. Energy involved is both kinetic energy due to motion and potential energy due to gain in height up the hill. Bridget is pedaling, hence he is transferring energy to the bicycle. Bridget is in motion along with the bicycle. Hence, both Bridget and Bicycle have kinetic energy and potential energy. We can say both have mechanical energy. Thus correct options are:

Bridget is transferring energy to the bicycle.

The bicycle is using energy to do work.

Bridget has kinetic energy.

The bicycle has potential energy.

The bicycle has mechanical energy.

3 0
3 years ago
Read 2 more answers
What is the required force to bring a 1550 kg vehicle moving at 32 m/s to a stop in a distance of 45 m?
Vilka [71]
The answer is 9 i think.
8 0
4 years ago
Choose the correct answer from the given alternatives
Black_prince [1.1K]

Answer:

d

Explanation:

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8 0
3 years ago
A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J
Vesna [10]

Answer:

16.9000000000000001 J

Explanation:

From the given information:

Let the initial kinetic energy from point A be K_A = 1.9000000000000001 J

and the final kinetic energy from point B be K_B = ???

The charge particle Q = 6 mC = 6 × 10⁻³ C

The change in the electric potential from point B to A;

i.e. V_B - V_A = -2.5 × 10³ V

According to the work-energy theorem:

-Q × ΔV = ΔK

-Q \times ( V_B - V_A) = (K_B - K_A)

-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)

15 = (K_B - 1.9000000000000001 \ J)

K_B = 15+ 1.9000000000000001 \ J

\mathbf{K_B =1 6.9000000000000001 \ J}

3 0
3 years ago
A 1400 kg car is traveling east on the highway at 31 m/s and collides into the rear of a slower moving pickup truck of 2400 kg,
zloy xaker [14]

Answer: 31 m/s due east

Explanation: this question can be solved using the law of conservation of linear momentum.

This law states that in a closed or isolated system, during collision, the vector sum of momentum before collision equals the vector sum of momentum after collision.

Momentum = mass × velocity

From our question, our parameters before collision are given below as

Mass of car = mc = 1400kg

Speed of car =vc = 31 m/s (due east)

Mass of truck = mt = 2400kg

Velocity of truck = vt = 25 m/s ( due east )

After collision

Velocity of car = ?

Velocity of truck = 34 m/s ( due east )

Vector sum of momentum before collision is given as

1400 (31) + 2400 (25) = 43400 + 60000 = 103400 kgm/s

After collision the truck is seen to move faster (v = 34 m/s) which implies that the car also moves due east .

1400 (v) + 2400(25) .... A positive value is between both momenta because they are in the same direction.

After collision, we have that

1400v + 60000

Vector sum of momentum before collision = vector sum of momentum after collision

103400 = 1400v + 60000

103400 - 60000 = 1400v

43400 = 1400v

v = 43400/ 1400

v = 31 m/s due east

4 0
4 years ago
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