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Drupady [299]
4 years ago
5

A ray of light strikes a smooth surface and is reflected. The angle of incidence is 35°. What can be predicted about the angle o

f reflection? Rays will bounce in all directions, so the angle of reflection will be random. Rays will bounce in all directions, so the angle of reflection will be less than 35°. Rays will follow the law of reflection, so the angle of reflection will be 35°. Rays will follow the law of reflection, so the
Physics
2 answers:
Scrat [10]4 years ago
8 0

Answer:

The answer is C:

Rays will follow the law of reflection, so the angle of reflection will be 35°.

Explanation:

I got a 100% on the Edge quiz with C as my answer for this question.

olga_2 [115]4 years ago
5 0
<span>Rays will follow the law of reflection, so the angle of reflection will be 35 degrees. If a</span> light ray strikes a smooth surface, the reflected ray will bounce off the surface with the same angle the ray hits the surface. In other words, the angle of incidence is the same of angle of reflection, which is 35 degrees. If the surface is not smooth, the reflected ray might diffuse in all directions.
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A tennis ball is dropped from 1.43 m above the
Rudiy27

Answer:

-5.29 m/s

Explanation:

Given:

y₀ = 1.43 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.43 m)

v = -5.29 m/s

4 0
4 years ago
Why did J.J. Thomson experiment with cathode ray tubes?
kkurt [141]
<span>Thomson's First Cathode Ray Experiment. Thomson had an inkling that the 'rays' emitted from the electron gun were inseparable from the latent charge, and decided to try and prove this by using a magnetic field. His first experiment was to build a cathode ray tube with a metal cylinder on the end.</span>
3 0
3 years ago
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A rocket with a mass of 2.0 Ã 106 kg is designed to take off from the surface of the earth by burning fuel and ejecting it with
julsineya [31]

thrust force getting from the burning of mass should balance the weight of the rocket

here thrust force is given as

F_t = v\frac{dm}{dt}

now by force balance we can say

mg = v \frac{dm}{dt}

now plug in all values in this

(2 \times 10^6)(9.8) = 3500 \times \frac{dm}{dt}

\frac{dm}{dt} = \frac{19.6 \times 10^6}{3500}

\frac{dm}{dt} = 5600 kg/s

so rate of mass burning per second will be 5600 kg per second in order to lift up the rocket

6 0
3 years ago
A heated piece of metal cools according to the function c(x) = (.5)^(x _ 11), where x is measured in hours. A device is added th
Agata [3.3K]
You just need to replace x with 5 in each function

.5^5 - 11
-5-3

.5 ^-6
-8


64 - 8 = 56 A Celcius

Hope this helps
3 0
3 years ago
Read 2 more answers
At some instant and location, the electric field associated with an electromagnetic wave in vacuum has the strength 57.5 V/m. Fi
n200080 [17]

Answer:

1.9167\times 10^{-7}\ T

2.9247498849\times 10^{-8}\ J/m^3

8.7742496547\ W/m^2

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

E = Electric field = 57.5 V/m

c = Speed of light = 3\times 10^8\ m/s

Magnetic field is given by

B=\dfrac{E}{c}\\\Rightarrow B=\dfrac{57.5}{3\times 10^8}\\\Rightarrow B=1.9167\times 10^{-7}\ T

The magnetic field strength is 1.9167\times 10^{-7}\ T

Energy density is given by

u=\dfrac{1}{2}\epsilon_0E^2+\dfrac{1}{2\mu_0}B^2\\\Rightarrow u=\dfrac{1}{2}\times 8.85\times 10^{-12}\times 57.5^2+\dfrac{1}{2\times 4\pi \times 10^{-7}}(1.9167\times 10^{-7})^2\\\Rightarrow u=2.9247498849\times 10^{-8}\ J/m^3

The energy density is 2.9247498849\times 10^{-8}\ J/m^3

Power flow per unit area is given by

\dfrac{P}{A}=uc\\\Rightarrow \dfrac{P}{A}=2.9247498849\times 10^{-8}\times 3\times 10^8\\\Rightarrow \dfrac{P}{A}=8.7742496547\ W/m^2

Power flow per unit area is 8.7742496547\ W/m^2

7 0
3 years ago
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