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Gwar [14]
3 years ago
7

A sonar wave is reflected from the ocean floor. For which angles of incidence do the wave’s angle of reflection equal its angle

of incidence?
Physics
1 answer:
nadya68 [22]3 years ago
7 0
After a thorough research, there exists the same question that has the following choices.

<span>(1) angles less than 45°, only
(2) an angle of 45°, only
(3) angles greater than 45°, only
<span>(4) all angles of incidence

The correct answer (4) all angles of incidence.</span></span>
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A diver shines a flashlight upward from beneath the water (n=1.33) at a 36.2° angle to the vertical. At what angle does the ligh
horrorfan [7]

Answer:

The flashlight leaves the water at an angle of 51.77°.

Explanation:

if n1 = 1.33 is the refractive index of water and ∅1 is the angle at which the flashlight shine beneath the water, and n2 = 1.0 is the refractive index of air and ∅2 is the angle the flashlight leaves the water.

Then, according to Snell's law :

n1×sin(∅1) = n2×sin(∅2)

sin(∅2) = n1×sin(∅1)/n2

           = (1.33)×sin(36.2)/(1.0)

           = 0.7855055×379

     ∅2 = 51.77°

Therefore, the flashlight leaves the water at an angle of 51.77°.

7 0
4 years ago
Answer all these and i'll mark brainliest.
Luda [366]
Formula for net force: force= mass x acceleration
6 0
3 years ago
A 2 kg watermelon is dropped from a 4 m tall roof. How fast will the watermelon be moving as it strikes the ground below?
Volgvan

Answer:

8.85m/s

Explanation:

v=√2gh

v=√2×9.8×4

v=8.85m/s

3 0
3 years ago
He blocked his man using a force of 23 N moving him backward 2 meters ,how much work did he do?
Paul [167]

Answer:

work done = force x distance

Explanation:

F = 23 N

D = 2m

W = 23 * 2 = 46 J

6 0
4 years ago
Two drag cars race. They line up at the starting line at rest. The winning car accelerates at a constant rate a and reaches the
Mila [183]

Answer:d=\frac{v^2}{8a}

Explanation:

Given

winning car accelerates with a and its final velocity is v

considering they both start from rest

time taken by winning car is

v=u+at

where u=initial velocity

a=acceleration

t=time

v=at

t=\frac{v}{a}

Now loosing car is accelerating with \frac{a}{4}

Distance traveled by loosing car in time t

s_1=ut+\frac{at^2}{2\cdot 4}

s_1=0+\frac{a}{8}\times (\frac{v}{a})^2

s_1=\frac{v^2}{8a}

Thus distance d traveled by loosing car is given by d=\frac{v^2}{8a}

5 0
4 years ago
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