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Arturiano [62]
3 years ago
11

A classic demonstration illustrating eddy currents is performed by dropping a permanent magnet inside a conducting cylinder. The

magnet does not go into free fall. Instead it reaches terminal velocity and can take a few seconds to drop a length of about a meter. Suppose the mass of the magnet is 70 g and width of 1.0 cm. It falls with a terminal velocity of 10 cm/s and the length of the pipe is 80 cm. The magnitude of the Joule heating from the eddy currents is approximately:________
Physics
1 answer:
kvv77 [185]3 years ago
4 0

Answer:

The correct solution is "0.69 N".

Explanation:

The given values are:

Mass of magnet,

m = 70 g

or,

   = 0.07 kg

Width,

= 1.0 cm

Velocity,

= 10 cm/s

Length of the pipe,

= 80 cm

Whenever the velocity is constant, then the net force which is acting on the magnet will be "0".

On the magnet,

The up-ward force will be:

⇒  F=mg

On substituting the values, we get

⇒      =0.07\times 9.8

⇒      =0.69 \ N

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The cochlea detects sound and is tonotopically organized

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6 0
2 years ago
Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains
OLEGan [10]

Answer:

C = 771.35 J/kg°C

Explanation:

Here, e consider the conservation of energy equation. The conservation of energy principle states that:

Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container

Since,

Heat Given or Absorbed by a material = m C ΔT

Therefore,

m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃

where,

m₁ = Mass of Metal Piece = 2.3 kg

C = Specific Heat of Metal = ?

ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C

m₂ = Mass of Metal Container = 3.8 kg

ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C

m₃ = Mass of Water = 20 kg

C₃ = Specific Heat of Water = 4200 J/kg°C

ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C

Therefore,

(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)

C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J

C = 252000 J/326.7 kg°C

<u>C = 771.35 J/kg°C</u>

5 0
3 years ago
Puzzle 3
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Answer:

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Explanation:

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4 0
3 years ago
Starting at t = 0 s , a horizontal net force F⃗ =( 0.285 N/s )ti^+(-0.460 N/s2 )t2j^ is applied to a box that has an initial mom
Irina-Kira [14]

Answer:

Explanation:

We know that Impulse = force x time

impulse = change in momentum

change in momentum = force x time

Force F = .285 t -.46t²

Since force is variable

change in momentum = ∫ F dt  where F is force

= ∫ .285ti - .46t²j dt

= .285 t² / 2i - .46 t³ / 3 j

When t = 1.9

change in momentum = .285 x 1.9² /2 i  -  .46 x 1.9³ / 3 j

= .514i - 1.05 j

final momentum

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7 0
3 years ago
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
guajiro [1.7K]

Answer:

f_{e} = 1.7 cm

Explanation:

The magnification of the compound microscope is given by the product of the magnification of each lens

        M = M₀ m_{e}

        M = - L/f₀  25/f_{e}

Where f₀ and f_{e} are the focal lengths of the lens and eyepiece, respectively, all values ​​in centimeters

In this exercise they give us the magnification (M = 400X), the focal length of the lens (f₀ = 0.6 cm), the distance of the tube (L = 16 cm), let's look for the focal length of the eyepiece (f_{e})

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Let's calculate

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        f_{e} = 1.67 cm

The minus sign in the magnification is because the image is inverted.

          f_{e} = 1.7 cm

6 0
3 years ago
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