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Oksi-84 [34.3K]
3 years ago
7

If you hold a bar magnet in each hand and bring your hands together, will the force be attractive or repulsive if the magnets ar

e held:
a) with the two north poles together?
b) with a north pole and south pole together?
Physics
1 answer:
Law Incorporation [45]3 years ago
5 0

a) The force is repulsive

b) The force is attractive

Explanation:

Every magnet has a magnetic field around it. It is possible to distinguish two different poles in the magnet, according to the direction of the magnetic field: in particular, the lines of the field go out from the North pole and go into the South Pole. When a magnet is broken, two new magnets are formed, each of them having its own north and south pole.

The force between two magnets can be either attractive or repulsive, depending on which poles are facing each other. We have the following situation:

  • The magnetic force between two like poles (north-north and south-south) is repulsive
  • The magnetic force between two opposite poles (north-south) is attractive

Therefore, we have the following situations in this problem:

a)

Here we are holding the two north poles together: since they are like poles, they repel each other, so the force in this case is repulsive

b)

Here we are holding a north pole and a south pole together: since they are opposite poles, they attract each other, so the force in this case is attractive

Learn more about magnetic fields:

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

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10 PTS! HELP!<br> If you have a mass of 47 kg on Earth, determine your mass on the moon.
Marta_Voda [28]

The mass would be the same

47kg on the moon as well

6 0
3 years ago
A mass of 250 N is on a piston of 2.0 m^2. What force is needed to lift this piston if the area of the second piston is 0.5 m^2?
prohojiy [21]

Answer:

<h3>62.5N</h3>

Explanation:

The pressure at one end of the piston is equal to the pressure on the second piston.

Pressure = Force/Area

F1/A1 = F2/A2

Given

F1 = 250N

A1 = 2.0m²

A2 = 0.5m²

F2 = ?

Substituting the given values in the formula;

250/2 = F2/0.5

cross multiply

250*0.5 = 2F2

125 = 2F2

F2 = 125/2

F2 = 62.5N

Hence the  force needed to lift this piston if the area of the second piston is 0.5 m^2 is 62.5N

8 0
3 years ago
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

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3 years ago
At what point does the external energy enter the system?
Phoenix [80]
The correct answer as the first one above !
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3 years ago
At t=0, a particle leaves the origin with a velocity of 9.0 m/s in the positive y direction and moves in the xy plane with a con
fiasKO [112]

Answer:

e.26m/s

Explanation:

Vf=Vi+at      (1)

Vf=9j+(2i-4j)t

X= X₀+at

now, in the i direction

15=O+2t or t=7.5 when x position is 15

Lets put that into the (1) equation, solve for Vf.

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Vf= 15i - 21j

Speed= \sqrt{xcomponent^2 + ycomponent^2}

Vf= 25.8 m/s

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3 years ago
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