Ask question

Ask question

All categories

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

You might be interested in

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

5 0

2 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

6 0

2 years ago

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

Maksim231197 [3]

<span> Corona </span>

<span>I. </span>Rare gaseous envelope

<span>II. </span>Irregular pearly glow

<span>III. </span>Total solar eclipse

<span>IV. </span>Surrounds darkened moon

<span>V. </span>Caused by diffraction

<span> Solar Prominence </span>

<span>I. </span>loop shape

<span>II. </span>large bright gaseous

<span>III. </span>luminous hydrogen gas

<span>IV. </span>rises above chromospheres

<span>V. </span>anchored to surface (sun’s surface)

<span> Solar Flare </span>

<span>I. </span>Intense High-energy

<span>II. </span>Causes electromagnetic disruptions

<span>III. </span>In solar atmosphere

<span>IV. </span>Varies in brightness

<span>V. </span>Emits radiation

<span> Sunspots </span>

<span>I. </span>Surface dark spots

<span>II. </span>Strong magnetic fields

<span>III. </span>Solar magnetic storms

<span>IV. </span> Visible through telescope

<span>V. </span>Cooler than photospheres

I did this in class I got 100%

7 0

2 years ago

ipn [44]

I need help to I have a test and I need help to

6 0

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)

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

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

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

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

6 0

2 years ago