PHYSICS COLLEGE

The index of refraction for red light in water is 1.331, and that for blue light is 1.340. If a ray of white light enters the water at an angle of incidence of 83.00o , what are the underwater angles of refraction for the blue and red components of the light

Answers

Answer 1

Answer:

Answer:

The underwater angles of refraction for the blue and red components of the light is 47.8° and 48.2°

Explanation:

Using the Snell's law

n1 * sin Ф1 = n2 sin Ф2

1 * sin 83 = n2 sin Ф2

Ф2 = $sin^(-1) ((1)/(n2) * sin 83)$

Red light

n2 = 1.331

Ф2 = $sin^(-1) ((1)/(1.331) * sin 83) = 48.2$ °

Blue light

n2 = 1.340

Ф2 = $sin^(-1) ((1)/(1.340) * sin 83) = 47.8$ °

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Chapter 38, Problem 001 Monochromatic light (that is, light of a single wavelength) is to be absorbed by a sheet of a certain material. Photon absorption will occur if the photon energy equals or exceeds 0.42 eV, the smallest amount of energy needed to dissociate a molecule of the material.

(a) What is the greatest wavelength of light that can be absorbed by the material?
(b) In what region of the electromagnetic spectrum is this wavelength located?

Answers

Answer:

a) $\lambda=2.95x10^(-6)m$

b) infrared region

Explanation:

Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".

Part a

Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:

$E=hf$

Where h is the Planck's Constant. By the other hand the know that $c=f\lambda$ and if we solve for f we have:

$f=(c)/(\lambda)$

If we replace the last equation into the E formula we got:

$E=h(c)/(\lambda)$

And if we solve for $\lambda$ we got:

$\lambda =(hc)/(E)$

Using the value of the constant $h=4.136x10^(-15) eVs$ we have this:

$\lambda=(4.136x10^(15)eVs (3x10^8 (m)/(s)))/(0.42eV)=2.95x10^(-6)m$

$\lambda=2.95x10^(-6)m$

Part b

If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of $10^(-6)m$

A train accelerates at -1.5 m/s2 for 10 seconds. If the train had an initialspeed of 32 m/s, what is its new speed?
A. 17 m/s
B. 15 m/s
C. 47 m/s
D. 32 m/s

Answers

Answer:

17 m/s

Explanation:

Using formula a = (v-u) /t

acceleration a = -1.5 m/s2

final velocity v = unknown

initial velocity u = 32 m/s

time t = 10s

-1.5 = (v-32)/10

-15 = v - 32

-15 + 32 = v

v = 17 m/s

2. An electrical heater 200mm long and 15mm in diameter is inserted into a drilled hole normal to the surface of a large block of material having a thermal conductivity of 5W/m·K. Estimate the temperature reached by the heater when dissipating 25 W with the surface of the block at a temperature of 35 °C.

Answers

Answer:

The final temperature is 50.8degrees celcius

Explanation:

Pls refer to attached handwritten document

Answer: 50.63° C

Explanation:

Given

Length of heater, L = 200 mm = 0.2 m

Diameter of heater, D = 15 mm = 0.015 m

Thermal conductivity, k = 5 W/m.K

Power of the heater, q = 25 W

Temperature of the block, = 35° C

T1 = T2 + (q/kS)

S can be gotten from the relationship

S = 2πL/In(4L/D)

On substituting we have

S = (2 * 3.142 * 0.2) / In (4 * 0.2 / 0.015)

S = 1.2568 / In 53.33

S = 1.2568 / 3.98

S = 0.32 m

Proceeding to substitute into the main equation, we have

T1 = T2 + (q/kS)

T1 = 35 + (25 / 5 * 0.32)

T1 = 35 + (25 / 1.6)

T1 = 35 + 15.625

T1 = 50.63° C

The spring constant, k, for a 22cm spring is 50N/m. A force is used to stretch the spring and when it is measured again it is 32cm long. Work out the size of this force

Answers

Answer:

Explanation:

Given parameters:

Original length = 22cm

Spring constant, K = 50N/m

New length = 32cm

Unknown

Force applied = ?

Solution:

The force applied on a spring can be derived using the expression below;

Force = KE

k is the spring constant

E is the extension

extension = new length - original length

extension = 32cm - 22cm = 10cm

convert the extension from cm to m;

100cm = 1m;

10cm will give 0.1m

So;

Force = 50N/m x 0.1m = 5N

Final answer:

To calculate the force used to stretch the spring, Hooke's Law is utilized, which leads to the conclusion that a force of 5 N was exerted to stretch the spring from its original length of 22 cm to a final length of 32 cm.

Explanation:

The force exerted by a spring is governed by Hooke's Law, which states that the force required to stretch or compress a spring by a certain distance is proportional to that distance. In this case, the spring constant, k, is given as 50 N/m and the spring is stretched from its original length of 22 cm to a final length of 32 cm. This represents a stretch, or displacement, of 10 cm (or 0.1 m when converted to the standard unit).

The force (F) can be calculated using Hooke's law: F = kx, where x is the displacement of the spring. Substituting the given values, the force amounts to F = (50 N/m) * (0.1 m) = 5 N. Therefore, the force used to stretch the spring to its final length of 32 cm is 5 N.

Learn more about Hooke's Law here:

brainly.com/question/32317230

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In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its speed is decreased linearly from 60 mph to 30 mph in 10 seconds. Calculate the theoretical maximum energy in kWh that can be recovered during this interval. Ignore all losses.

Answers

Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

m = 4000 / 2.20462 = 1814.37 kg

Initial velocity $V_(i)$ = 60 mph = 26.8224 m/s

Final velocity $V_(f)$ = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = $(1)/(2)$ m( $V_(i)$ ² - $V_(f)$ ² )

we substitute

Δk = $(1)/(2)$ ×1814.37( (26.8224)² - (13.4112)² )

Δk = $(1)/(2)$ × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Suppose that a particle accelerator is used to move two beams of particles in opposite directions. In a particular region, electrons move to the right at 6020 m/s and protons move to the left at 1681 m/s. The particles are evenly spaced with 0.0476 m between electrons and 0.0662 m between protons. Assuming that there are no collisions and that the interactions between the particles are negligible, what is the magnitude of the average current in this region