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3 years ago

What W is the amount of force with which gravity is pulling down on something

Physics

1 answer:

stira [4]3 years ago

4 0

Answer: Weight– the amount of gravity acting on (pulling down on) an object (or mass).

Gravity– a natural force that pulls objects downward.

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A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the

Rama09 [41]

1) $1.11\cdot 10^{-7} J$

The capacitance of a parallel-plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

$r=\frac{2.0 cm}{2}=1.0 cm=0.01 m$

so the area is

$A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2$

While the separation between the plates is

$d=0.50 mm=5\cdot 10^{-4} m$

So the capacitance is

$C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F$

And now we can find the energy stored,which is given by:

$U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J$

2) 0.71 J/m^3

The magnitude of the electric field is given by

$E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m$

and the energy density of the electric field is given by

$u=\frac{1}{2}\epsilon_0 E^2$

and using

$E=4\cdot 10^5 V/m$ , we find

$u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3$

7 0

4 years ago

A machinist turns the power on to a grinding wheel, at rest, at time t = 0 s. The wheel accelerates uniformly for 10 s and reach

natka813 [3]

The total number of revolutions made by the wheel is closest to is 28.2 revolutions. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

5 0

3 years ago

Read 2 more answers

How is it technically correct to say that a car making a u-turn can have a constant speed but cannot have a constant velocity?

saw5 [17]

During the "U" part of the turn, the car would follow an approximately circular path, and if it's moving at a constant speed, it would have to accelerate toward the center of the circle in order to change its direction.