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Vedmedyk [2.9K]
3 years ago
12

The angle of incidence must equal the angle of reflection is always less than the angle of reflection is always greater than the

angle of reflection may be greater than, less than, or equal to the angle of reflection
Physics
1 answer:
saw5 [17]3 years ago
6 0

Answer: The angle of incidence is not always equal to the angle of reflection.

Explanation:

The angle of incidence might not be equal to the angle of reflection. It depends of the type of surface in consideration. If the surface is smooth, the incident ray will reflect out at the same angle the incident ray makes with surface. This is not the same for a rough or irregular surface.

For an irregular surface, the angle of incident is not equal to the angle if reflection because the reflected ray always reflects at different angles to the horizontal.

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A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is
Soloha48 [4]

Answer: V=5839.051m/s  

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;:

T=1.26(10)^{4}s is the period of the satellite

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=5.98(10)^{24}kg is the mass of the Earth

a  is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity V is given by:

V=\sqrt{\frac{GM}{a}}   (2)

Now, from (1) we can find a, in order to substitute this value in (2):

a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}   (3)

a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}   (4)

a=11705845.57m   (5)

Substituting (5) in (2):

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}   (6)

V=5839.051m/s   (7)  This is the speed at which the satellite travels

6 0
3 years ago
Help me please. <br> A<br> B<br> C<br> D
Y_Kistochka [10]

I will go to school tomorrow .....is this present tense or past tense or future tense

5 0
2 years ago
Read 2 more answers
A 0.140-kg baseball is dropped from rest from a height of 1.8 m above the ground. It rebounds to a height of 1.4 m. What change
aliina [53]

Answer:

\Delta p=-1.56\ kg-m/s

Explanation:

It is given that,

Mass of the baseball, m = 0.14 kg

It is dropped form a height of 1.8 m above the ground. Let u is the velocity when it hits the ground. Using the conservation of energy as :

u=\sqrt{2gh}

h = 1.8 m  

u=\sqrt{2\times 9.8\times 1.8}

u = 5.93 m/s

Let v is the speed of the ball when it rebounds. Again using the conservation of energy to find it :

v=\sqrt{2gh'}

h' = 1.4 m  

v=-\sqrt{2\times 9.8\times 1.4}

v = -5.23 m/s

The change in the momentum of the ball is given by :

\Delta p=m(v-u)

\Delta p=0.14(-5.23-5.93)

\Delta p=-1.56\ kg-m/s

So, the change in the ball's momentum occurs when the ball hits the ground is 1.56 kg-m/s. Hence, this is the required solution.

4 0
3 years ago
A car of mass 150 kg is riding down at constant velocity. What is the normal force?
xz_007 [3.2K]

Answer:

1500N

Explanation:

Normal force = mg - F sin theta

constant velocity means acceleration = 0

F= ma = 150× 0 = 0N

thus;

normal force = mg = 150 × 10 = 1500N

4 0
3 years ago
Read 2 more answers
The simplest form of a traveling electromagnetic wave is a plane wave. For a wave traveling in the x direction whose electric fi
musickatia [10]

Answer:

amplitudes

Explanation:

In everyday physics we define the amplitude of a wave as the maximum (this can also be called the highest)displacement or distance moved by a point on a given vibrating body or wave as measured from its equilibrium position. The key idea in defining the amplitude of a wave motion is the idea of a 'maximum displacement from the position of equilibrium'.

Given the equations;

E= EoSin(kx - ωt)y

B= Bosin(kx- ωt)z

Both Eo and Bo refer to the maximum displacement of the electric and magnetic field components of the electromagnetic wave. This maximum displacement is known as the amplitude of the electric and magnetic components of the electromagnetic wave.

4 0
3 years ago
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