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soldier1979 [14.2K]

3 years ago

9

A circular loop of radius 0.10 m is rotating in a uniform external magnetic field of 0.20 T. Find the magnetic flux through the

loop due to the external field when the plane of the loop and the magnetic field vector are parellel.

1 answer:

laiz [17]3 years ago

4 0

Answer:

The magnetic flux through the loop is 0.006284 m².T

Explanation:

Given;

radius of circular loop, r = 0.1 m

magnetic field strength, B = 0.2 T

Magnetic flux through the loop due to the external field is given as;

Φ = BAcosθ

where;

Φ is magnetic flux

B is the magnetic field strength

A is the area of the loop

θ is the angle of inclination of between the plane of the loop and external magnetic field

A = πr² = π(0.1)² = 0.03142 m²

θ = 0, since they are parallel

Φ = BAcosθ

Φ = 0.2 x 0.03142 x cos0

Φ = 0.006284 m².T

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Answer:

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Explanation:

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3 years ago

A cube of ice at an initial temperature of -15.00°C weighing 12.5 g total is placed in 85.0 g of water at an initial temperature

Leona [35]

<u>Answer:</u> The specific heat of ice is 2.11 J/g°C

<u>Explanation:</u>

When ice is mixed with water, the amount of heat released by water will be equal to the amount of heat absorbed by ice.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of ice = 12.5 g

$m_2$ = mass of water = 85.0 g

$T_{final}$ = final temperature = 22.24°C

$T_1$ = initial temperature of ice = -15.00°C

$T_2$ = initial temperature of water = 25.00°C

$c_1$ = specific heat of ice = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$12.5\times c_1\times (22.24-(-15))=-[85.0\times 4.186\times (22.24-25)]$

$c_1=2.11J/g^oC$

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4 years ago

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sweet-ann [11.9K]

Answer:

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where,

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h = height = 1.16 m

Vf = Final Velocity of Ball = ?

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Therefore, using these values in the equation, we get:

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Latent heat $L=33.5\times 10^4 J/kg$

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Heat require to convert water at $T=0^{\circ}$ to ice at $T=0^{\circ}$

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Total heat $Q=Q_1+Q_2+Q_3$

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