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3 years ago

5

A 0.19 F capacitor is desired. What area must the plates have if they are to be separated by a 2.0 mm air gap? a= ?? m^2

1 answer:

Aliun [14]3 years ago

4 0

.125
Hope this answer helps you

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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 $7.5\times10^{-7}\ Watt$ .

6 0

3 years ago

Read 2 more answers

A sled is moving at a constant speed down a surface inclined at 45 degrees with the horizontal and travels 30meter in 4 seconds.

Vika [28.1K]

Constant speed along the inclined surface = 30 m / 4 s = 7.5 m/s

Vertical speed = inclined speed * sin(45) = 7.5 *√2 / 2 = 5.3 m/s

Answer: 5.3 m/s

8 0

3 years ago

Help me please I can't get the final step

inna [77]

Answer:

$\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}$

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

$\displaystyle A=\frac{3}{2}B^mC^n$

and the dimensions of each variable is:

$A=L^2T^2$

$B=LT^{-1}$

$C=LT^2$

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

$\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n$

Operating:

$L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)$

$L^2T^2=L^{m+m}T^{-m+2n}$

Equating the exponents:

$m+n=2$

$-m+2n=2$

Adding both equations:

$3n=4$

Solving:

$n=4/3$

$m=2-4/3=2/3$

Answer:

$\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}$

6 0

2 years ago

Consult Interactive Solution 10.37 to explore a model for solving this problem. A spring is compressed by 0.0647 m and is used t

padilas [110]

Answer:

$\omega=32.14\ rad/s$

Explanation:

Given that,

The compression in the spring, x = 0.0647 m

Speed of the object, v = 2.08 m/s

To find,

Angular frequency of the object.

Solution,

We know that the elation between the amplitude and the angular frequency in SHM is given by :

$v=\omega\times A$

A is the amplitude

In case of spring the compression in the spring is equal to its amplitude

$\omega=\dfrac{v}{A}$

$\omega=\dfrac{2.08\ m/s}{0.0647\ m}$

$\omega=32.14\ rad/s$

So, the angular frequency of the spring is 32.14 rad/s.

4 0

3 years ago

Control rods in a nuclear reactor are used to:

kiruha [24]

B.absorb neutrons to prevent chain reactions which become uncontrollable

7 0

3 years ago

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