List two applications of electrostatics.

1 answer:

4 0

Atmospheric electricity and storms,electric current in a vacuum,spark discharge,electrostatic control filters and industrial electrostatic separation <- those are just a few

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An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu

Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

$\alpha = \frac{\omega_f - \omega_i}{t}$

where we have

$\alpha = -80.0 rad/s^2$ is the angular acceleration (negative since the rotor is slowing down)

$\omega_f$ is the final angular speed

$\omega_i = 2000 rad/s$ is the initial angular speed

t = 10.0 s is the time interval

Solving for $\omega_f$ , we find the final angular speed after 10.0 s:

$\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s$

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

$\alpha = \frac{\omega_f - \omega_i}{t}$

If we re-arrange it for t, we get:

$t = \frac{\omega_f - \omega_i}{\alpha}$

where here we have

$\omega_i = 2000 rad/s$ is the initial angular speed

$\omega_f=0$ is the final angular speed

$\alpha = -80.0 rad/s^2$ is the angular acceleration

Solving the equation,

$t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s$

6 0

3 years ago

A subway train accelerates from rest at one station at a rate of 1.30 m/s^2 for half of the distance to the next station, then d

Minchanka [31]

This problems a perfect application for this acceleration formula:

   Distance = (1/2) (acceleration) (time)² .

During the speeding-up half:   1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half: 1,600 meters = (1/2) (1.3 m/s²) T²

Pick either half, and divide each side by 0.65 m/s²:

   T² = (1600 m) / (0.65 m/s²)

   T = square root of (1600 / 0.65) seconds

Time for the total trip between the stations is double that time.

   T = 2 √(1600/0.65) = <em>99.2 seconds</em> (rounded)

7 0

3 years ago

If an object has a mass of 20 grams and a volume of 40 cm3, what is its density in g/cm3?

sattari [20]

Density = Mass/Volume
So, given mass = 20 g and volume = 40 cm^3
By substituting in above equation, Density = 20/40 = 0.5 g/cm^3
Hope it helps.

3 0

3 years ago

Two identical charges are separated by a distance d. If the distance between them is increased to 3d, what will happen to the fo

belka [17]

A-It will be one-ninth the original force

3 0

3 years ago