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olganol [36]
3 years ago
10

A 2.98 nF parallel-plate capacitor is charged to an initial potential difference of 49 V and then isolated. The dielectric mater

ial between the plates has a dielectric constant of 3.1. What is the potential difference of the capacitor after the dielectric is withdrawn? Answer in units of V.
Physics
2 answers:
I am Lyosha [343]3 years ago
8 0

Answer:

Potential difference will be 151.9 volt  

Explanation:

We have given capacitance of the capacitor C=2.98nF=2.98\times 10^{-9}F

Voltage V = 49 Volt

Dielectric constant K = 3.1

We have to find the potential difference

We know that when a dielectric medium is introduced then p[otential difference is increases by k times

As the dielectric constant k = 3.1

So potential difference will be = 3.1×49 = 151.9 volt

docker41 [41]3 years ago
6 0

Answer:151.9 V

Explanation:

Given

Capacitance (C)=2.98 nF

Potential difference=49 V

dielectric strength k=3.1

Charge remains same after the removal of dielectric

thusQ_1=Q_2

Q_1=2.98\times 49

Now dielectric is removed so capacitance decreases to \frac{C}{k}

Q_2=\frac{2.98}{3.1}\times V

V=49\times 3.1=151.9 V

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3 years ago
A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o
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The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

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If the incline is 20.4 m long, that means the block has a starting height of

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The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

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where n is the magnitude of the normal force, so that

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Then the block slides a distance of

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Natasha_Volkova [10]

Answer:

421.83 m.

Explanation:

The following data were obtained from the question:

Height (h) = 396.9 m

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

First, we shall determine the time taken for the ball to get to the ground.

This can be calculated by doing the following:

t = √(2h/g)

Acceleration due to gravity (g) = 9.8 m/s²

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t = √(2h/g)

t = √(2 x 396.9 / 9.8)

t = √81

t = 9 secs.

Therefore, it took 9 secs fir the ball to get to the ground.

Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:

Time (t) = 9 secs.

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

s = ut

s = 46.87 x 9

s = 421.83 m

Therefore, the horizontal distance travelled by the ball is 421.83 m

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Answer:

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