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2 years ago

When a dielectric slab is inserted between the plates of one of the two identical capacitors

Physics

2 answers:

tatiyna2 years ago

4 0

Answer:(A-P,S;B-P,S;C-Q,S;D-P,S)

Solution

(A)→P,S,(B)→P,S,(C)→Q,S,(D)→P,S.

Explanation:

Akimi4 [234]2 years ago

3 0

This could be like 711 u know they be selling hot dogs they good try them

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If you have lifted 10 pounds 2 feet, you have done ___ foot-pounds of work.

nasty-shy [4]

Work = Force x distance
(10 pounds)(2 feet)
Work = 20 foot-pounds of work

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3 years ago

An experimental apparatus has two parallel horizontal metal rails separated by 1.0 m. A 3.0 Ω resistor is connected from the lef

Blizzard [7]

Answer:

The induced current and the power dissipated through the resistor are 0.5 mA and $7.5\times10^{-7}\ Watt$ .

Explanation:

Given that,

Distance = 1.0 m

Resistance = 3.0 Ω

Speed = 35 m/s

Angle = 53°

Magnetic field $B=5.0\times10^{-5}\ T$

(a). We need to calculate the induced emf

Using formula of emf

$E = Blv\sin\theta$

Where, B = magnetic field

l = length

v = velocity

Put the value into the formula

$E=5.0\times10^{-5}\times1.0\times35\sin53^{\circ}$

$E=1.398\times10^{-3}\ V$

We need to calculate the induced current

$E =IR$

$I=\dfrac{E}{R}$

Put the value into the formula

$I=\dfrac{1.398\times10^{-3}}{3.0}$

$I=0.5\ mA$

(b). We need to calculate the power dissipated through the resistor

Using formula of power

$P=I^2 R$

Put the value into the formula

$P=(0.5\times10^{-3})^2\times3.0$

$P=7.5\times10^{-7}\ Watt$

Hence, The induced current and the power dissipated through the resistor are 0.5 mA and $7.5\times10^{-7}\ Watt$ .

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3 years ago

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Sedaia [141]

1. Proton
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4. electron
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3 years ago

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Citrus2011 [14]

Answer:

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Explanation:

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2 years ago

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Aloiza [94]

Answer:

Force and displacement.

Explanation:

Work done is positive when we push table and it move in the direction of applied force.

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2 years ago