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Sholpan [36]
3 years ago
15

How did the iron filing patterns show attractive forces between magnetic poles?

Physics
2 answers:
nasty-shy [4]3 years ago
4 0

When the magnet is placed on the glass, it is attracted to the iron filings. The pattern of the iron filings shows that the lines of force that make up the magnetic field of the magnet. Also, The lines of force of north and south poles attract each other whereas those of two north poles fend off each other.

julia-pushkina [17]3 years ago
3 0

Answer:

When the magnet was placed on the glass, it attracted the iron filings. The pattern of the iron filings shows the lines of force that make up the magnetic field of the magnet. ... The lines of force of north and south poles attract each other whereas those of two north poles repel each other.

Explanation:

Hope it's answer you plz mark as breinlist !!!

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The Earth's atmosphere is kept close to Earth by gravity. Just as the Earth's gravity keeps you firmly on the ground, it also acts on the gases of the Earth's atmosphere, holding it in place and allowing us to breathe.
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If you push on a ball with a force of 200 N, what force does the ball push back?
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100N divide 2 by 200 it is 100
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3 years ago
Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add
jok3333 [9.3K]

Answer:

When x = 2.8 cm, B_{x1} = 0.0265 T

When x = 5.5 cm, B_{x2} = 0.0209 T

when x = 7.3 cm, B_{x3} = 0.0169 T

When x = 11.0 cm, B_{x4} = 0.0103 T

Explanation:

According to Biot-Savart law,

B_{x} = \frac{N \mu_{o}IR^{2}  }{2(x^{2} +R^{2}  )^{3/2} }\\.......................(1)

R = 11.0 cm = 0.11 m

I = 17.0 A

N = 300 turns

\mu_{o}  = 4\pi  * 10^{-7} N/A^{2}

When x₁ = 2.8 cm = 0.028 m

B_{x1} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.028^{2} +0.11^{2}  )^{3/2} }\\B_{x1} = 0.0265 T

When x₂ = 5.5cm = 0.055 m

B_{x2} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.055^{2} +0.11^{2}  )^{3/2} }\\B_{x2} = 0.0209 T

When x₃ = 7.3 cm = 0.073 m

B_{x3} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.073^{2} +0.11^{2}  )^{3/2} }\\B_{x3} = 0.0169 T

When X₄ = 11.0 cm = 0.11 m

B_{x4} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.11^{2} +0.11^{2}  )^{3/2} }\\B_{x4} = 0.0103 T

4 0
4 years ago
4. Solid ‘X’ turns into a liquid at 80 0C and into a gas at 140 0C. Describe the changes in the
Paladinen [302]

Answer:

Explanation:

The melting point of the solid is 80°C

        Vapor point of the liquid is 140°C

What happens to particles of X when heated from 70°C to 85°C?

  • Firstly, there would be a phase change from solid to liquid.
  • Below the melting point, a substance will exist as a solid.
  • With increase in thermal energy inputted by heat, as the temperature climbs above the melting point, it changes to the liquid.
  • When the solid begins to heat up, the particles of X starts vibrating about their fixed point.
  • At they melting point, they break lose and flow to form a liquid.
  • The particles will have more kinetic energy.
5 0
3 years ago
Consider massive gliders that slide friction-free along a horizontal air track. Glider A has a mass of 1 kg, a speed of 1 m/s, a
mamaluj [8]

Answer:

0.167m/s

Explanation:

According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.

Given momentum = Maas × velocity.

Momentum of glider A = 1kg×1m/s

Momentum of glider = 1kgm/s

Momentum of glider B = 5kg × 0m/s

The initial velocity of glider B is zero since it is at rest.

Momentum of glider B = 0kgm/s

Momentum of the bodies after collision = (mA+mB)v where;

mA and mB are the masses of the gliders

v is their common velocity after collision.

Momentum = (1+5)v

Momentum after collision = 6v

According to the law of conservation of momentum;

1kgm/s + 0kgm/s = 6v

1 =6v

V =1/6m/s

Their speed after collision will be 0.167m/s

6 0
3 years ago
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