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3 years ago

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A light ray in air hits piece of quartz at a 39.0 deg angle. Inside the quartz, the light slows down to 2.06*10^8 m/s. At what a

ngle does the light travel in the quarts? (Hint: find n.)
(Unit=deg)

1 answer:

Naddika [18.5K]3 years ago

3 0

Answer:

Angle of light at which it will travel in the quartz is given as

$\theta = 25.6^o$

Explanation:

As we know that the refractive index of a medium is the ratio of speed of light in air and speed of light in the medium

So we will have

$\mu = \frac{c}{v}$

here we have

$\mu = \frac{3 \times 10^8}{2.06 \times 10^8}$

$\mu = 1.456$

now by Snell's law we have

$n_1 sin\theta_1 = n_2 sin\theta_2$

$1 sin39 = 1.456 sin\theta$

$\theta = 25.6^o$

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A spring with a spring constant of 165 N/m is attached to a 2.0 kg mass and set into motion.

marin [14]

Explanation:

We have,

Spring constant of the spring, k = 165 N/m

Mass, m = 2 kg

It is required to find the period of the mass-spring system. For the spring mass system, the period is given by :

$T=2\pi \sqrt{\dfrac{m}{k} }\\\\T=2\pi \sqrt{\dfrac{2}{165} }\\\\T=0.69\ s$

The frequency of vibration is reciprocal of its time period. So,

$f=\dfrac{1}{T}\\\\f=\dfrac{1}{0.69}\\\\f=1.44\ Hz$

So, the period of the mass-spring system is 0.69 s and frequency is 1.44 Hz.

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4 years ago

Ainat [17]

Explanation:

Hey there!!

Here, Given is,

Efficiency = 75%

VR = no. of pulleys = 5

Now,

$effeciency = (\frac{ma}{vr} ) \times 100\%$

$75\% = \frac{ma}{5} \times 100\%$

100% ma = 75%×5

$ma = \frac{75\% \times5}{100\%}$

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Therefore, MA is 3.75.

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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3 years ago

Read 2 more answers

The alignment of atoms where the north poles face one way and the south poles face the opposite way is also known as which of th

Pepsi [2]

The answer is B. Domain
When the atoms point to the same direction and aligned with one another, It will form a magnetic material in which magnetization is in a uniform direction
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3 years ago

Solenoid 2 has twice the diameter, twice the length, and twice as many turns as solenoid 1. How does the field B2 at the center

klemol [59]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is option 3

Explanation:

From the question we are told that

The diameter of solenoid 1 is $d_1$

The length of solenoid 1 is $L_1$

The number of turns of solenoid is $N_1$

The diameter of solenoid 2 is $d_2 = 2d_1$

The length of solenoid 2 is $L_2 = 2L_1$

The number of turns of solenoid 2 is $N_2 = 2 N_1$

Generally the magnetic in a solenoid is mathematically represented as

$B = \frac{\mu_o * N * I }{L}$

From this equation we see that

$B \ \alpha \ \frac{N}{L}$

$B = C \frac{N}{L}$

Here C stands for constant

=> $C = \frac{B * \frac{L}{N}$

=> $\frac{B_1 * \frac{L_1}{N_1} = \frac{B_2 * \frac{L_2}{N_2}$

=> $\frac{B_1}{B_2 } = \frac{N_1 L _2}{ N_2L_1}$

=> $\frac{B_1}{B_2 } = \frac{N_1 * (2 L_1)}{ (2 N_2)L_1}$

=> $\frac{B_1}{B_2 } = 1$

=> $B_2 = B_1$

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4 years ago

A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a

Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

$Z_{R}$ = R

R = resistance

$Z_{L}$ = jωL

ω = voltage source angular frequency, L = inductance

$Z_{C}$ = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

$Z_{R}$ = 217Ω

$Z_{L}$ = j(220)(0.875) = j192.5Ω

$Z_{C}$ = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$ + $Z_{C}$

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

$v_{R}$ = I× $Z_{R}$

$v_{L}$ = I× $Z_{L}$

$v_{C}$ = I× $Z_{C}$

Given values:

I = (0.0569∠1.147+220t)A

$Z_{R}$ = 217Ω = (217∠0)Ω

$Z_{L}$ = j192.5Ω = (192.5∠π/2)Ω

$Z_{C}$ = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

$v_{R}$ = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

$v_{L}$ = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

$v_{C}$ = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

$v_{R}$ = (12.35∠1.147+220t)V

$v_{L}$ = (10.95∠2.718+220t)V

$v_{C}$ = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

$v_{R}$ = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

$v_{L}$ = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

$v_{C}$ = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0

3 years ago