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2 years ago

Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated by insulation.

Physics

1 answer:

andrew-mc [135]2 years ago

7 0

Answer:

The eddy currents will be very small.

Explanation:

An object made of several thin conducting layers separated by insulation may not be affected by magnetic damping because the eddy current produced in each layer due to induction will be very small and the opposing magnetic flux produced by the eddy currents will be very small.

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A scientist studies how sunlight affects a person's skin. Which statement is a

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If you spend more time in the sun, your skin will become drier.

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A material through which electronics do NOT easily flow is

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Resistor. The answer is a resistor. They do not transmit electricity.

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2 years ago

The magnitude of a force vector is 89.6 newtons (N). The x component of this vector is directed along the +x axis and has a magn

insens350 [35]

Answer:

(a) θ = 33.86°

(b) Ay = 49.92 N

Explanation:

You have that the magnitude of a vector is A = 89.6 N

The x component of such a vector is Ax = 74.4 N

(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:

$A_x=Acos\theta$ (1)

Ax: x component of vector A

A: magnitude of vector A

θ: angle between vector A and the x axis

You solve the equation (1) for θ, by using the inverse of cosine function:

$\theta=cos^{-1}(\frac{A_x}{A})=cos{-1}(\frac{74.4N}{89.6N})\\\\\theta=33.86\°$

the angle between the A vector and the x axis is 33.86°

(b) The y component of the vector is given by:

$A_y=Asin\theta\\\\A_y=(89.6N)sin(33.86\°)=49.92N$

the y comonent of the vecor is Ay = 49.92 N

3 0

2 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

3 years ago

A parallel-plate capacitor has square plates that are 8.00cm on each side and 4.20mm apart. The space between the plates is comp

Sergeu [11.5K]

Answer:

$U=1.29\times 10^{-7}\ J$

Explanation:

Given that

a= 8 cm (square)

A= a ² = 64 cm²

d= 4.2 mm

d₁= 2.1 mm ,K₁= 4.7

d₂=2.1 mm , K₂=2.6

We know that capacitance given as

$C_1=\dfrac{K_1\varepsilon _oA}{d_1}$

$C_1=\dfrac{4.7\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}$

$C_1=1.26\times 10^{-10}\ F$

$C_2=\dfrac{K_2\varepsilon _oA}{d_2}$

$C_2=\dfrac{2.6\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}$

$C_2=0.701\times 10^{-10}\ F$

Net capacitance

$C=\dfrac{C_1C_2}{C_1+C_2}$

$C=\dfrac{1.26\times 10^{-10}\times 0.701\times 10^{-10}}{1.26\times 10^{-10}+0.701\times 10^{-10}}\ F$

$C=4.5\times 10^{-11}\ F$

We know that stored energy given as

$U=\dfrac{CV^2}{2}$

V= 76 V

$U=\dfrac{4.5\times 10^{-11}\times 76^2}{2}\ J$

$U=1.29\times 10^{-7}\ J$

3 0

2 years ago