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3 years ago

Moving a magnet inside a coil of wire will induce a voltage in the coil. How can the voltage in the coil be increased?

2 answers:

5 0

As the magnet is moved inside a coil of wire, the number of lines of magnetic field passing through the coil changes. Faraday stated that : it is the change in the number of field lines passing through the the coil of wire that induces emf in the loop. Specifically, it is the rate of change in the number of magnetic field lines passing through the loop that determines the induced emf. There is a term called magnetic flux same as electric flux, this magnetic flux can be a measure of the number of field lines passing through a surface. It is given by ( Φ=ΣB. dA. Where B is magnetic field and dA is small elementary area). The induced emf is given by (ξ = dΦ/dt). This equation states that THE MAGNITUDE OF THE INDUCED CURRENT IN A CIRCUIT IS EQUAL TO THE RATE AT WHICH THE MAGNETIC FLUX THROUGH THE CIRCUIT IS CHANGING WITH TIME. So more rapid you move the coil, more will be the change in flux and hence more emf will be produced. So option D is the correct answer. I hope this long description will help you out.

Lapatulllka [165]3 years ago

5 0

_Award brainliest if helped!
The voltage produced is proportional to the rate of change of flux (magnetic field)
D. Move the magnet inside the coil of wire more rapidly.

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Different asteroids reflect different percentages of the light falling on them. This is due to the fact that they have different

Georgia [21]

Due to the Composition, different asteroids reflect different percentages of the light falling on them.

What are Asteroids:

Asteroids are small, rocky objects that orbit the Sun. Although asteroids orbit the Sun like planets, they are much smaller than planets.
Asteroids are generally made up of rocky material, metals and their size are large in comparison to comets. Asteroid belt is found between Jupiter and Mars.

Composition of Asteroids:

Most of the asteroids in the Main Belt are made of rock and stone.

The remaining asteroids are made up of a mix of these, along with carbon-rich materials. Some of the more distant asteroids tend to contain more ices.

We determine reflectivity of asteroids by comparing the brightness of light in the visible spectrum to the brightness of light in the infrared spectrum. The light shining from asteroids is reflected sunlight.

Hence we can say that,

Due to the Composition, different asteroids reflect different percentages of the light falling on them.

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1 year ago

Which items are matter? Pick multiple

AleksAgata [21]

Answer:

maybe phone cause matter is a solid liquid or gad

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3 years ago

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn

KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

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3 years ago

A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$

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3 years ago

As a mercury atom absorbs a photon of

Allisa [31]

The energy absorbed by photon is 1.24 eV.
This is the perfect answer.

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2 years ago