A teacher walks 5m north of his desk, then we walks 6m south.

A narrow beam of light from a laser travels through air (n = 1.00) and strikes the surface of the water (n = 1.33) in a lake at

Natalka [10]

Answer:

A) d = 11.8m

B) d = 4.293 m

Explanation:

A) We are told that the angle of incidence;θ_i = 70°.

Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;

tan 70° = d/4.3m

Where d is the distance from point B at which the laser beam would strike the lakebottom.

So,d = 4.3*tan70

d = 11.8m

B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)

So,

n1*sinθ_i = n2*sinθ_r

Thus; sinθ_r = (n1*sinθ_i)/n2

sinθ_r = (1 * sin70)/1.33

sinθ_r = 0.7065

θ_r = sin^(-1)0.7065

θ_r = 44.95°

Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;

d = 4.3 tan44.95

d = 4.293 m

4 0

3 years ago

g The speed of light in a vacuum is larger for short wavelength electromagnetic waves. larger for high frequency electromagnetic

Orlov [11]

Answer:

A constant value everywhere in the universe.

Explanation:

The speed of light in a vacuum is a constant value. It is not affected by change in frequency or wavelength of the light.

Mathematically the speed of light is given as:

c = λf

where λ = wavelength and f - frequency

The speed of light is the constant of proportionality between frequency and wavelength. In order words, wavelength and frequency are inversely proportional. As the wavelength increases, frequency decreases and vice versa.

While the change in wavelength and frequency of light affect the energy of the light, its speed is a constant value as long as the medium is a vacuum.

The speed of light is also not dependent on the manner with which the light wave is moving.

4 0

3 years ago

If i = 1.70 A of current flows through the loop and the loop experiences a torque of magnitude 0.0760 N ⋅ m , what are the lengt

Alik [6]

Answer:

Length of the sides of the square loop is given by

s = √[(τ)/(NIB sin θ)]

Explanation:

The torque, τ, experienced by a square loop of area, A, with N number of turns around the loop and current of I flowing in the wire, with a magnetic field presence, B, and the plane of the loop tilted at angle θ to the x-axis, is given by

τ = (N)(I)(A)(B) sin θ

If everything else is given, the length of a side of the square loop, s, can be obtained from its Area, A.

A = s²

τ = (N)(I)(A)(B) sin θ

A = (τ)/(NIB sin θ)

s² = (τ)/(NIB sin θ)

s = √[(τ)/(NIB sin θ)]

In this question, τ = 0.076 N.m, I = 1.70 A

But we still need the following to obtain a numerical value for the length of a side of the square loop.

N = number of turnsof wire around the loop

B = magnetic field strength

θ = angle to which the plane of the loop is tilted, measured with respect to the x-axis.

6 0

3 years ago

Read 2 more answers

For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103

Alona [7]

Answer:

a) P = 44850 N

b) $\delta l =0.254\ mm$

Explanation:

Given:

Cross-section area of the specimen, A = 130 mm² = 0.00013 m²

stress, σ = 345 MPa = 345 × 10⁶ Pa

Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa

Initial length, L = 76 mm = 0.076 m

a) The stress is given as:

$\sigma=\frac{\textup{Load}}{\textup{Area}}$

on substituting the values, we get

$345\times10^6=\frac{\textup{Load}}{0.00013}$

or

Load, P = 44850 N

Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>

b)The deformation ( $\delta l$ ) due to an axial load is given as:

$\delta l =\frac{PL}{AE}$

on substituting the values, we get

$\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}$

or

$\delta l =0.254\ mm$

3 0

3 years ago

Object A and Object B have the same volume of 10 cm³. Object A has a mass of 20 grams. Object B has a mass of 50 grams. Which ob

aniked [119]

Answer:

Object B has greater density

desity A=20/10=2 g cm^-3 . density B=50/10=5 g cm^-3

the object that has greater mass has the greater density because the volume of the those two objects are same

3 0

3 years ago