When this current is closed which way does the current flow

Answers

Answer 1

Answer: Well, Godess, that's not a simple question, and it doesn't have
a simple answer.

When the switch is closed . . .

"Conventional current" flows out of the ' + ' of the battery, through R₁ ,
then through R₂ , then through R₃ . It piles up on the right-hand side of
the capacitor (C). It repels the ' + ' charges on the left side of 'C', and
those flow into the ' - ' side of the battery. So the flow of current through
this series circuit is completely clockwise, around toward the right.

That's the way the first experimenters pictured it, that's the way we still
handle it on paper, and that's the way our ammeters display it.

BUT . . .

About 100 years after we thought that we completely understand electricity,
we discovered that the little tiny things that really move through a wire, and
really carry the electric charge, are the electrons, and they carry NEGATIVE
charge. This turned our whole picture upside down.

But we never changed the picture ! We still do all of our work in terms of
'conventional current'. But the PHYSICAL current ... the actual motion of
charge in the wire ... is all exactly the other way around.

In your drawing ... When the switch is closed, electrons flow out of the
' - ' terminal on the bottom of the battery, and pile up on the left plate of
the 'C'. They repel electrons off of the right-side of 'C', and those then
flow through R₃ , then through R₂ , then through R₁ , and finally into the
' + ' terminal on top of the battery.

Those are the directions of 'conventional' current and 'physical' current
in all circuits.

In the circuit of YOUR picture that you attached, there's more to the story:

Battery current can't flow through a capacitor. Current flows only until
charges are piled up on the two sides of 'C' facing each other, and then
it stops.

Wait a few seconds after you close the switch in the picture, and there is
no longer any current in the loop.

To be very specific and technical about it . . .

-- The instant you close the switch, the current is

(battery voltage) / (R₁ + R₂ + R₃) amperes

but it immediately starts to decrease.

-- Every (C)/((R₁ + R₂ + R₃) seconds after that, the current is

e⁻¹ = about 36.8 %

less than it was that same amount of time ago.

Now, are you glad you asked ?

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Do magnets work through glass?

Answers

Yes but the thicker the glass the stronger the magnet has to be

In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is approximately 2.20 3 106 m/s. Find (a) the force acting on the electron as it revolves in a circular orbit of radius 0.529 3 10210 m and (b) the centripetal acceleration of the electron

Answers

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A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of begin quantity omega times t end quantity, where V is the maximum voltage, omega is the angular frequency, and t is the time. The current supplied by this source that flows through this resistor is described with the function i of t is equal to I times cosine of begin quantity omega times t end quantity, where I is the maximum current. What is the average power supplied by this AC source?

Answers

Answer:

In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:

$p(t)=v(t)i(t)$

So the average power is:

$P=(1)/(T)\intop_(0)^(T)p(t)dt$

But:

$v(t)=v_(m)cos(\omega t) \n \n i(t)=i_(m)cos(\omega t)$

So:

$P=(1)/(T)\intop_(0)^(T)v_(m)cos(\omega t)i_(m)cos(\omega t)dt \n \n P=(v_(m)i_(m))/(T)\intop_(0)^(T)cos^(2)(\omega t)dt \n \n But: cos^(2)(\omega t)=(1+cos(2\omega t))/(2)$

$P=(v_(m)i_(m))/(T)\intop_(0)^(T)((1+cos(2\omega t))/(2) )dt \n\nP=(v_(m)i_(m))/(T)\intop_(0)^(T)[(1)/(2)+(cos(2\omega t))/(2)]dt \n\nP=(v_(m)i_(m))/(T)[(1)/(2)(t)\right|_0^T +(sin(2\omega t))/(4\omega) \right|_0^T] \n \n P=(v_(m)i_(m))/(2T)[(t)\right|_0^T +(sin(2\omega t))/(2\omega) \right|_0^T] \n \n P=(v_(m)i_(m))/(2)$

In terms of RMS values:

$V_(RMS)=V=(v_(m))/(√(2)) \n \n I_(RMS)=I=(i_(m))/(√(2)) \n \n Then: \n \n P=VI$

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Answers

I have the exact same question, any chance you figured it out since you posted this?

A ski jumper competing for an Olympic gold medal wants to jump a horizontal distance of 135 meters. Thetakeoff point of the ski jump is at a height of 25 meters. With what horizontal speed must he leave the jump?
a. What do you know?
b. What do you need to solve for?
c. What equation(s) will you use?
d. What is the solution to this problem?

Answers

A. We know that the height (h) = 25 m, and the distance (x) = 135 m.

B. We need to find the time flight first (t), then we can find easily the horizontal velocity ( $v_(x)$ . )

C. Because this is a projectile motions, so horizontal velocity is constant, but the vertical velocity is change because of the gravitational acceleration.
So we can use : $h = (1)/(2) g t^(2)$ ,

Note : Time that will be exist is a time from a land to the takeoff point, that can be called as the time flight.
Then we use, $x = v_(x) t$ , to find the horizontal velocity.

D. We know that :
$h = (1)/(2) g t^(2)$ , so by insert the numbers, and assune that the (g = 10 m/ $s^(2)$ ), we get that ( t = $√(5)$ . )
After that by using $x = v_(x) t$ , we can get that $v_(x) = 27 √(5)$

Forces that are unequal in size but opposite in direction will cause movement.A) true
B)false

Answers

Answer:

true

Explanation:

if they were balance then it would not move

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