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4 years ago

What are the three law of reflection?

Physics

1 answer:

Radda [10]4 years ago

6 0

Answer:

<h2>The three law of reflection are:-</h2>

The incident ray, the reflected ray, and normal ray all lie in same plane.
The angle of reflection is always equal to the angle of incidence.
Incident ray and reflected ray are on different sides of the normal.

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A transverse wave on a long horizontal rope with a

frosja888 [35]

Answer:

2 seconds

Explanation:

The frequency of a wave is related to its wavelength and speed by the equation

$f=\frac{v}{\lambda}$

where

f is the frequency

v is the speed of the wave

$\lambda$ is the wavelength

For the wave in this problem,

v = 2 m/s

$\lambda=8 m$

So the frequency is

$f=\frac{2}{8}=0.25 Hz$

The period of a wave is equal to the reciprocal of the frequency, so for this wave:

$T=\frac{1}{f}=\frac{1}{0.25}=4 s$

This means that the wave takes 4 seconds to complete one full cycle.

Therefore, the time taken for the wave to go from a point with displacement +A to a point with displacement -A is half the period, therefore for this wave:

$t=\frac{T}{2}=\frac{4}{2}=2 s$

4 0

3 years ago

A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac

RoseWind [281]

Answer:

a) $\Delta U_{g} = 12.945\,J$ , b) $\Delta U_{k} = 12.945\,J$ , c) $k = 2930.059\,\frac{N}{m}$

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

$\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)$

$\Delta U_{g} = 12.945\,J$

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

$\Delta U_{k} = 12.945\,J$

c) The spring constant of the gun is:

$\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}$

$k = \frac{2\cdot \Delta U_{k}}{x^{2}}$

$k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}$

$k = 2930.059\,\frac{N}{m}$

4 0

3 years ago

During the period that the moon’s phases are changing from new to full, the moon is _____. waxing exhibiting retrograde motion w

Sindrei [870]

The answer that best fits the blank is the term WAXING. The moon is waxing whenever it reaches to the period that its phases are transitioning from new to full. The answer is the first option. This is when it is more that half is illuminated. Hope this helps.

3 0

3 years ago

Read 2 more answers

Gravity on the moon is about 1/6 th the gravity felt on the earth. This is because A) the moon is so far away from the earth. B)

ankoles [38]

<span>So we want to know why is there a difference between the force of gravity on the Moon and the force of gravity of the Earth. So the gravitational force between two objects depends on the masses of both objects. That can be seen from Newtons universal law of gravity. F=G*m1*m2*(1/r^2). So lets say we are holding an object of mass m=1kg on a height r=1m on the Moon and we are holding the same object on the Earth also on the same height of r=1m. The Gravitational force on the Earth will be Fg=G*M*m*(r^2) where M is the mass of the Earth. The force between the moon and that object will be Fg=G*n*m*(r^2), where n is the mass of the moon. Since mass of the Moon is much smaller than mass of the Earth, The gravitational force between the Moon and that body will be almost 6 times smaller than the gravitational force between the Earth and that body. So the correct answer is B. </span>

4 0

3 years ago

Read 2 more answers

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

What are the three law of reflection?​

What are the three law of reflection?