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Troyanec [42]
2 years ago
9

Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of 190

mm^2. How much charge must be transferred from one plate to the other if 1.1 nJ of energy are to be stored in the plates? ( ε0 = 8.85 × 10^-12 C2/N · m^2)
Physics
1 answer:
julia-pushkina [17]2 years ago
8 0

Answer:

5.5\cdot 10^{-11} C

Explanation:

The capacitance of the parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

where

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=190 mm^2 = 190 \cdot 10^{-6} m^2 is the area of the plates

d=1.2 mm = 0.0012 m is the separation between the plates

Substituting,

C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F

The energy stored in the capacitor is given by

U=\frac{Q^2}{2C}

Since we know the energy

U=1.1 nJ = 1.1 \cdot 10^{-9} J

we can re-arrange the formula to find the charge, Q:

Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C

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Explanation:

In diffraction pattern we know that the distance between two consecutive minimum position is maximum

So here position of minimum is given as

asin\theta = N\lambda

now we have central maximum is of maximum width while all other have width of decreasing order.

While when we use Young's double slit pattern then we can say that position of all maximum and minimum intensity on the screen will be at same distance.

so here we have

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Which condition is necessary for total internal reflection?
AleksAgata [21]

Answer:

The index of refraction of the first medium must be higher than the index of refraction of the second medium

Explanation:

Snell's law describes the behaviour of light at the boundary between two mediums:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n1 and n2 are the index of refraction of the two mediums

\theta_1, \theta_2 are the angle between the direction of the light ray and the normal to the interface

We can rewrite the condition as:

sin \theta_2 = \frac{n_1}{n_2} sin \theta_1

Let's assume now that the light is travelling in the first medium with a very large angle with respect to the normal to the surface, i.e. \theta_1 = 90^{\circ}, so that sin \theta_1=1. In this case, we have

sin \theta_2 = \frac{n_1}{n_2}

We notice that if n_1 > n_2, the ratio on the right is larger than 1, and so the term sin \theta_2 should be also larger than 1: but this is not possible of course, since the sine function is always less than 1. Therefore, in this case total internal reflection occurs, because no refracted ray is produced.


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3 years ago
A particle moves in the xy plane, starting from the origin at t=0 with initial velocity having an x component of 20 m/s and a y
777dan777 [17]

Answer:

a. \vec{v} = \vec{v_x} + \vec{v_y} = (20 + 4t)\hat{x} - 15 \hat{y}

b. 25m/s

Explanation:

Let t be the time.

The velocity in the x direction that is subjected to acceleration of 4 m/s2

v_x = 20 + 4t m/s

The velocity in the y direction

v_y = -15 m/s

The total velocity at any time, which is the combination of both x and y vectors of velocity

\vec{v} = \vec{v_x} + \vec{v_y} = (20 + 4t)\hat{x} - 15 \hat{y}

b) at t = 5

v^2 = v_x^2 + v_y^2 = (20 + 4t)^2 + 15^2 = (20 + 4*5)^2 + 15^2 = 400^2 + 225 = 625

v = \sqrt{625} = 25 m/s

5 0
3 years ago
A guitar string is 90 cm long and has a mass of 3.4g . The distance from the bridge to the support post is L=62cm, and the strin
Drupady [299]

Answer:

f = 293.38 Hz

1 st overtone = 586.76 Hz

2 nd overtone = 880.14 Hz

Explanation:

Mass of the string, m =3.4 g = 0.0034 kg

Length of the string, l = 90 cm = 0.90 m

Distance  from the bridge, L = 62 cm = 0.62 m

Tension in the string, T = 500 N

Fundamental frequency is given by:

f=\frac{\sqrt{Tl/m}}{2L}\\ \Rightarrow f=\frac{\sqrt{500\times0.90/0.0034}}{2\times0.62}=293.38 Hz

First overtone = 2 × f = 2 × 293.38 =586.76 Hz

Second Overtone = 3×f = 3 × 293.38 = 880.14 Hz

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