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Gelneren [198K]
3 years ago
12

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500

Physics
1 answer:
padilas [110]3 years ago
5 0

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

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maks197457 [2]

Heavy crate sits at rest on the floor of a warehouse. you push on the crate with a force of 400 N, and it doesn't budge. The magnitude of the friction force on the crate in Newton is 400N

This is due to Friction force, which is defined as the resisting force that acts on a body when it is at rest (Static friction) or when it is in motion (Kinetic friction).

When a force is applied on a stationary body, the force of static friction starts to act on the body which prevents any relative motion between the object and surface. The magnitude of friction increases up to μsN, where μs is the coefficient of static friction. As the crate didn't budge, it means the amount of force applied was less than μsN. Hence the force applied was canceled by an equal and opposite amount of frictional force which was equal to 400N.

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8 0
2 years ago
A 1 800-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.60 m before coming into contact
N76 [4]

Answer:

F = 614913.88 N

Explanation:

We are given;

Mass of pile driver; m = 1800 kg

Height of fall of pole driver; h = 4.6 m

Depth driven into beam; d = 13.6 cm = 0.136 m

Now, from energy equations and applying to this question, we can write that;

Workdone = Change in potential energy

Formula for workdone is; W = F × d

While the average potential energy here is; W = mg(h + d)

Thus;

Fd = mg(h + d)

Where F is the average force exerted by the beam on the pile driver while in bringing it to rest.

Making F the subject, we have;

F = mg(h + d)/d

F = 1800 × 9.81 × (4.6 + 0.136)/0.136

F = 614913.88 N

7 0
3 years ago
I need to lift a 2000kg car, 1.798m and the joules required is 35240.8. Converted to watt (W = 35240.8/5 (s)) I got 7048.16 W. I
marusya05 [52]
This is a very interesting problem ... mainly because it's different from
the usual questions in the Physics neighborhood.

I can discuss it with you, but maybe not quite give you a final answer
with the information you've given in the question.

I agree with all of your calculations so far ... the total energy required,
and the power implied if the lift has to happen in 5 seconds.

First of all, let's talk about power.  I'm assuming that your battery is
a "car" battery, and I'm guessing you measured the battery voltage
while the car was running.  Turn off the car, and you're likely to read
something more like 13 to 13.8 volts.
But that's not important right now.  What I'm looking for is the CURRENT
that your application would require, and then to look around and see whether
a car battery would be capable of delivering it.

   Power = (volts) x (current)

   7,050 W  =  (14 volts) x (current)

   Current = (7,050 watts / 14 volts) =  503 Amperes. 

That kind of current knocks the wind out of me.  I've never seen
that kind of number outside of a power distribution yard.
BUT ... I also know that the current demand from a car battery during
starting is enormous, so I'd better look around online and try to find out
what a car battery is actually capable of.

I picked a manufacturer's name that I'd heard of, then picked their
recommended battery for a monster 2003-model car, and looked at
the specs for the battery.

The spec I looked at was the 'CCA' ... cold cranking Amps.
That's the current the battery is guaranteed to deliver for 30 seconds,
at a temperature of 0°F, without dropping below 12 volts.

This battery that I saw is rated  803 Amps  CCA !

OK.  Let's back up a little bit.  I'm pretty sure the battery you have
is a nominal "12-volt" battery.  Let's say you use to start lifting the lift. 
As the lift lifts, the battery voltage sags.  What is the required current
if the battery immediately droops to 12V and stays there, while delivering
7,050 watts continuously ?

          Power = (volts) x (current)

          7,050 W = (12 V) x (current)

            Current = (7,050 W / 12 V)  =  588 Amps . 

Amazingly, we may be in the ball park.
If the battery you have is rated by the manufacturer for 600 Amps
CCA (0°F) or CA (32°F), then the battery can deliver the current
you need.
BUT ... you can't conduct that kind of current through ear-bud wire,
or house wiring wire.  I'm not even so sure of jumper-cables. 
You need thick, no-nonsense cable, AND connections with a lot of
area ... No alligator clips.  Shiny nuts and bolts with no crud on them.

Now ... I still want to check the matter of the total energy.
I'm sure you're OK, because the CCA and CA specifications talk about
30 seconds of cranking, and you're only talking about 5 seconds of lifting.
But I still want to see the total energy requirement compared to the typical
battery specification ... 'AH' ... ampere-hours.

You're talking about 35,000 joules

                          = 35,000 watt-seconds

                         =  35,000 volt-amp-seconds.

               (35,000 volt-amp-sec) x (1 hour/3600 sec) / (12 volt)               

           =  (35,000 x 1) / (3600 x 12)  volt-amp-sec-hour / sec-volt

           =    0.81 Amp-Hour  .

That's an absurdly small depletion from your car battery.
But just because it's only  810 mAh, don't get the idea that you can
do it with a few rechargeable AA batteries out of your camera.
You still need those 600 cranking amps.  That would be a dead short
for a stack of camera batteries, and they would shrivel up and die.

Have I helped you at all ?
5 0
3 years ago
Newton has Three Laws of Motion. Seat belts are used to help keep the passengers remaining in their seats during an accident. If
Jobisdone [24]
That is the 1st Law
Your welcome and PLEASE RATE TO KEEP HELPING YOU
5 0
3 years ago
1 point
Kazeer [188]
P = m v
250 = m5

250 = 5m

m = 50 Kg

Option C
5 0
3 years ago
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