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Anastaziya [24]
3 years ago
8

The electric potential inside a parallel-plate capacitor __________. the electric potential inside a parallel-plate capacitor __

________. decreases linearly from the negative to the positive plate decreases inversely with the square of the distance from the negative plate is constant increases linearly from the negative to the positive plate decreases inversely with distance from the negative plate
Physics
1 answer:
suter [353]3 years ago
3 0
<span>The electrical potential inside a parallel-plate capacitor will increase linearly from the negative to the positive plate. If the charge is moving in the direction is naturally would move, the electric potential energy is said to decrease; if it moves in the opposite direction, it is said to increase. This is akin to lifting an object further away from the surface of the earth. As it is lifted higher, the electrical potential energy increases.</span>
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A transverse traveling wave on a cord is represented by, where D and x are in meters and t is in seconds.
yarga [219]

Answer with Explanation:

We are given that a transverse travelling wave on a cord is represented by

D=0.51sin(6.1x+76t)

Where D and x in meters

t(in seconds)

a.General equation of transverse wave

y=Asin(kx+\omega t)

By comparing we get

A=0.51

k=6.1

k=\frac{2\pi}{\lambda}

Wavelength,\lambda=\frac{2\pi}{k}=\frac{2\pi}{6.1}=1.03

b.\omega=76

Frequency,f=\frac{\omega}{2\pi}=\frac{76}{2\pi}=12.096Hz

c.Velocity,v=f\lambda=12.096\times 1.03=12.46m/s

Direction:Towards negative x- axis

d.Amplitude,A=0.51 m

e.Maximum speed,v_{max}=A\omega=0.51\times 76=38.76 m/s

Minimum speed,v_{min}=0

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What is the magnitude of a vector that has the following components: x = 32 m y = -59 m
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Answer:

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Explanation:

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With the switch open, roughly what must be the resistance of the resistor on the right for the current out of the battery to be
jeyben [28]

Question:

Assumptions

Voltage of battery = 24 V

Resistance on the right,20Ω parallel to 10 Ω resistor

Answer:

For the current out of the battery to be the same as when the switch was opened with the switch closed, the resistance on the resistor on the right must be approximately 20/3 Ω

Explanation:

We note that the switch in the assumption is on the same line as the 20 Ω resistor.

With a voltage of 24 V, and the switch closed, we have;

Total resistance =  \frac{1}{\frac{1}{10} +\frac{1}{20} } = \frac{20}{3} \Omega

Current out of voltage, I  = Voltage/(total resistance)

= 24 ÷ 20/3 = 24 × 3/20 = 18/5 A

Therefore, with the switch opened, we get

Resistance on the right = Initial total resistance =  \frac{20}{3} \Omega

Therefore, with the switch opened, the resistance on the resistor on the right must be approximately equal to the resultant resistance of the two resistances in parallel.

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