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Troyanec [42]
3 years ago
13

A flat coil having 160 turns, each with an area of 0.20 m 2, is placed with the plane of its area perpendicular to a magnetic fi

eld of 0.40 T. The magnetic field changes uniformly from 0.40 T in the + x direction to 0.40 T in the - x direction in 2.0 s. If the resistance of the coil is 16 Ω, at what rate is power generated in the coil during this change?
Physics
1 answer:
S_A_V [24]3 years ago
6 0

Answer:

10.2 Watt

Explanation:

N  = number of turns in flat coil = 160

A  = area = 0.20 m²

B₀= initial magnetic field = 0.40 T

B  = final magnetic field = - 0.40 T

Change in magnetic field is given as

ΔB = B - B₀ = - 0.40 - 0.40 = - 0.80 T

t  = time taken for the magnetic field to change = 2.0 s

Induced emf is given as

E = \frac{- N A \Delta B}{t}

E = \frac{- (160) (0.20) (- 0.80)}{2}

E = 12.8 volts

R = Resistance of the coil = 16 Ω

Power is given as

P = \frac{E^{2}}{R}

P = \frac{(12.8)^{2}}{16}

P = 10.2 Watt

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Explanation:

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3 years ago
An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10
Alexeev081 [22]

Answer:

a) c=1822.3214\ J.kg^{-1}.K^{-1}

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

  • mass of aluminium, m_a=0.1\ kg
  • mass of water, m_w=0.25\ kg
  • initial temperature of the system, T_i=10^{\circ}C
  • mass of copper block, m_c=0.1\ kg
  • temperature of copper block, T_c=50^{\circ}C
  • mass of the other block, m=0.07\ kg
  • temperature of the other block, T=100^{\circ}C
  • final equilibrium temperature, T_f=20^{\circ}C

We have,

specific heat of aluminium, c_a=910\ J.kg^{-1}.K^{-1}

specific heat of copper, c_c=390\ J.kg^{-1}.K^{-1}

specific heat of water, c_w=4186\ J.kg^{-1}.K^{-1}

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)

Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)

Q_{in}=11375\ J

The heat released by the blocks when dipped into water:

Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)

where

c= specific heat of the unknown material

For the conservation of energy : Q_{in}=Q_{out}

so,

11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)

c=1822.3214\ J.kg^{-1}.K^{-1}

b)

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c)

The material is peat, possibly.

d)

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

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Answer:

Option (C)

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Thus, the correct answer is option (C).

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Answer:

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