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Evgen [1.6K]
3 years ago
8

Two people of different masses sit on a seesaw. m1, the mass of person 1, is 93 kg, m2 is 44 kg, d1 = 0.8 m, and d2 = 1.2 m. the

mass of the board is negligible. (a) what is the magnitude of the torque about the pivot (location
a.due to the gravitational force on person 1? |τa,1| = n · m
Physics
1 answer:
Vikentia [17]3 years ago
7 0
Torque = Force * Distance

Force = Weight = m1*g = 93*9.81 = 912.33 N
Distance from the  pivot to person 1, d1 = 0.8 m

Therefore,

Torque = 912.33*0.8 = 729.864 N.m
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What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500
padilas [110]

Answer:

I = 1.21x10^-5 A

Explanation:

You are missing the first part of the problem. This is an example, but it will give you the idea of how to solve yours with your data.

The first part is like this:

<em>A      4.0 cm  diameter parallel plate capacitor has a  0.44 m  m    gap. What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V/s?</em>

Now with this, we can solve the problem.

In order to do this, we need to use the following expression:

q = CV (1)

Where:

C: Capacitance of a parellel capacitor (in Faraday)

q: charge of plate or capacitor (In coulombs)

V: voltage in Volts.

However, we need is the current, and we have data of potential difference, so, all we have to do is divide the expression between time so:

q/t = CV/t

And the current is q/t, thus:

I = C * V/t (2)

And finally, Capacitance C with two plates of area A separated by a distance d is:

C = Eo*A/d (3)

Where:

Eo = constant equals to 8.85x10^-12 F/m.

A = Area of the plate, in this case, πr²

d = gap of the capacitor.

Let's calculate first the Capacitance using equation (3):

C = 8.85x10^-12 * π * (0.04/2)² / 0.00046 = 2.42x10^-11 F

Now, it's time to use equation (2) and solve for I:

I = 2.42x10^-11 * 500,000

I = 1.21x10^-5 A

5 0
3 years ago
Help ASAP
EastWind [94]

Answer:

Elastic potential energy, E=2.35\times 10^{-8}\ J

Explanation:

Charge, q=9.4\times 10^{-10}\ C

Potential, V = 50 V

It is required to find the electric potential energy in a capacitor stored in it. The formula of the electric potential energy in a capacitor is given by :

E=\dfrac{1}{2}qV\\\\E=\dfrac{1}{2}\times 9.4\times 10^{-10}\times 50\\\\E=2.35\times 10^{-8}\ J

So, the electric potential energy stored in the capacitor is 2.35\times 10^{-8}\ J

8 0
3 years ago
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