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3 years ago

What is the speed of light in (a) crown glass n= 1.52, (b) Lucite n = 1.51, and (c) ethyl alcohol n = 1.36?

Physics

1 answer:

mote1985 [20]3 years ago

4 0

Answer:

$(a)\\1.97\times 10^8 m/s\\(b)\\1.99\times 10^8 m/s \\(c)\\2.21 \times 10^8 m/s$

Explanation:

According to the definition of refractive index, the refractive index of the material is defined as the ratio of speed of light in vacuum to the speed of light in medium.

(a) refractive index, n = 1.52

$n=\frac {c}{v}\\\\1.52 = \frac{3\times 10^8}{v}\\\\v = 1.97\times 10^8 m/s$

(b) refractive index, n = 1.51

$n=\frac {c}{v}\\\\1.51 = \frac{3\times 10^8}{v}\\\\v = 1.99\times 10^8 m/s$

$n=\frac {c}{v}\\\\1.36 = \frac{3\times 10^8}{v}\\\\v = 2.21\times 10^8 m/s$

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Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope

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a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

= -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

= 26,754 (0.816)

= 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c. What is the work done by the gravitational force on the sled?

By using the formula:

Gravity work = mgdsinθ

= 97.5 kg * 9.8 m/s² * 28 m * sin 60

= 23,170 J

23,170 J the work, in joules done by the gravitational force on the sled .

D. What is the total work done?

By adding all the values

work done = -1337.3 + 21,835 + 23,170

= 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

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2 years ago

In the lab, you shoot an electron towards the south. As it moves through a magnetic field, you observe the electron curving upwa

Marta_Voda [28]

Answer:

Explanation:

An electron in a magnetic field always experience a force that tends to change its direction of motion through the magnetic field. According to Lorentz left hand rule (which is the opposite of Lorentz right hand rule for a positive charge), the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.

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3 years ago

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$

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3 years ago