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3 years ago

6

A parallel-plate capacitor is constructed from two aluminum foils of 1 square centimeter area each placedon both sides of a rubb

er square of the same size. The rubber dielectric is 2.5 mm thick, hasr2.5, andbreakdown field strength of 25 megavolts per meter. Find the voltage rating of the capacitor using a safetyfactor of 10.

1 answer:

Svet_ta [14]3 years ago

4 0

Answer:

The voltage will be 0.0125V

Explanation:

See the picture attached

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A voltage amplifier with an input resistance of 40k ohms, an output resistance of 100 ohms, and a gain of 300 V/V is connected b

AfilCa [17]

Answer:

89.45 v/v

Explanation:

Let's take the data:

First draw the amplifier circuit.

After the circuit, the voltage division rule can be used to compute the parameters:

The input section is computed like this: $\frac{R_{in} }{(R_{sig} + R_{in} } (v_{sig})$

The output section is computed like this $R_{L}/ (R_out} + R_{in} )$

The product A $V_{in} V_{out}$ gives

A $V_{in} V_{out}$ = A× $\frac{R_{in} }{(R_{sig} + R_{in} } (v_{sig})$ × $R_{L}/ (R_out} + R_{in} )$

Computing gives output voltage = 89.45 v/v

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3 years ago

A 7.0kg object rests on a horizontal frictionless surface. What is the magnitude of the horizontal

sukhopar [10]

Answer:

16.1 N

Explanation:

From the question,

F = ma.............................. Equation 1

Where F = horizontal force, m = mass of the object, a = acceleration .

Given: m = 7.0 kg, a = 2.3 m/s²

Substitute this values into equation 1

F = (7.0×2.3)

F = 16.1 N.

Hence the magnitude of the horizontal force is 16.1 N

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2 years ago

I will fan and give 2 medals!!!!!!

Maksim231197 [3]

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2 years ago

ipn [44]

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3 years ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

a)

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

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2 years ago