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Rama09 [41]
3 years ago
8

Which phenomenon occurs when light falls on a smooth mirror? . . A). Specular reflection. B) diffusion. C) refraction . D) diffu

sed refraction
Physics
2 answers:
gizmo_the_mogwai [7]3 years ago
8 0
Hey there!!!
your answer is going to be a, specular reflection.
hope this helps!!!
SCORPION-xisa [38]3 years ago
3 0
Reflection over a smooth surfaces such as mirrors or water leads to a type of reflection called specular reflection. Reflection over  rough surfaces such as clothing, paper etc. leads to a type of reflection called diffuse reflection<span>.
</span>Therefore,

Specular Reflection occurs when light falls on a smooth mirror.
So the right answer is :
<span>A). Specular reflection.</span>
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Marysya12 [62]

Answer: 0.28 m

Deceleration, a= -60 g

where g is the acceleration due to gravity=9.8 m/s^2

So, a= -60\times 9.8 m/s^2=-588 m/s^2

Using the equation of motion, we need to find the initial velocity,

v-u=at

final velocity, v=0

time, t= 31 ms=31\times 10^{-3}s

0-u=-588m/s^2\times 31\times 10^{-3} s=18.23 m/s

now, using third equation of motion,

s=ut+\frac{1}{2}at^2

we can find out the distance traveled by the person before coming to complete stop.

s=18.23m/s\times 31\times 10^{-3}s+\frac{1}{2}(-588 m/s^2) \times (31\times 10^{-3} s)^2=0.56 m-0.28 m=0.28 m


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3 years ago
A boy throws rocks with an initial velocity of 12m/s [down] from a 20 m bridge into a river. Consider the river to be at a heigh
stira [4]

Answer:

Please find attached the Velocity-Time graph, the Displacement-Time graph and the combined Velocity/Displacement-Time graph, created with Microsoft Excel

Explanation:

The given parameters are;

The initial velocity with which the boy throws the rock, u = 12 m/s

The direction in which he throws the rock = Down

The height from which the rock was thrown, h = 20 m

The height at which the river is located = 0 m

The kinematic equation of motion, of the rock can be given as follows;

h = u·t + 1/2·g·t²

Where;

t = The time of motion of the rock

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the rock is thrown = 20 m

Substituting the known values into given equation, we get;

20 = 12·t + 1/2·9.8·t² = 12·t + 4.9·t²

4.9·t² + 12·t - 20 = 0

t = (-12 ± √(12² - 4×4.9×(-20)))/(2 × 4.9) = (-12 ± √(242))/(9.8)

t ≈ -3.587 seconds or t ≈ 1.138 seconds

Attached please find the Velocity-Time graph, the Displacement-Time graph and the combined Velocity/Displacement-Time graph, created with Microsoft Excel.

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