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3 years ago

How do you increase capacitance in a capacitor?

1 answer:

4 0

If you want to increase the Capacitance of Parallele Plate Capacitor then increase the surface area, reduce the separation between the plate and use a dielectric material in between the plate which have higher dielectric breakdown strength.

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A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri

vodka [1.7K]

We know that tangential acceleration is related with radius and angular acceleration according the following equation:
at = r * aa
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)
So the radius is r = d/2 = 1.2/2 = 0.6 m
Then at = 0.6 * 5 = 3 m/s2
Tangential acceleration of a point on the flywheel rim is 3 m/s2

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3 years ago

So i want to start a dance academy. what do i do

timama [110]

Answer:

get students and a license

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3 years ago

Read 2 more answers

A bowling ball has a mass of 5 kg. A student rolls the bowling ball one time at 1 m/s and a second time and 2 m/s. Compare the m

sasho [114]

The answer is 5kg m/s with the second momentum being 10 kg m/s. i took the test and it was right. hope this helps yalls

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3 years ago

Jenny wanted to test whether worms prefer light or darkness best. After

myrzilka [38]

Answer:

True

Explanation:

An independent variable is the variable that is changed or controlled in a scientific experiment to test the effects on the dependent variable. You are changing light or dark

3 0

3 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

3 years ago