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jok3333 [9.3K]
4 years ago
7

I need this fast thank you❣️

Physics
2 answers:
mel-nik [20]4 years ago
7 0
The rule for the sequence is: You add 4/6 each time
5/6 +4/6 = 1 1/2
1 1/2 + 4/6 = 2 1/6
2 1/6 +4/6 = 2 5/6
muminat4 years ago
3 0
The rule is that 4/6 is added each time, and when adding or subtracting a fraction the denominator stays the same
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I'm confused with this​ I need this ASAP
Gnoma [55]
I’m pretty sure it’s a
6 0
3 years ago
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
Two astronauts are floating close to each other in space. Can they talk to each other without using any special device? plsss he
storchak [24]

Answer:

no they can't talk to each other bcoz of the lack of atmosphere.

Explanation:

l hope it helps you

5 0
3 years ago
Please dont delete this question......
Juli2301 [7.4K]

Answer:

c

Explanation:

8 0
3 years ago
Read 2 more answers
What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction
Sergio039 [100]

Answer:

The frequency is 302.05 Hz.

Explanation:

Given that,

Speed = 18.0 m/s

Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .

We need to calculate the frequency

Using formula of frequency

f'=f(\dfrac{v+v_{p}}{v-v_{s}})

Where, f = frequency

v = speed of sound

v_{p} = speed of passenger

v_{s} = speed of source

Put the value into the formula

f'=262\times(\dfrac{344+18}{344-30})

f'=302.05\ Hz

Hence, The frequency is 302.05 Hz.

7 0
3 years ago
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