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Strike441 [17]
4 years ago
15

An angle of refraction is the angle between the refracted ray and the

Physics
2 answers:
ivolga24 [154]4 years ago
5 0
<span>An angle of refraction is the angle between the refracted ray and the "Normal ray"

Hope this helps!</span>
erastova [34]4 years ago
5 0

Answer: The correct answer is the Normal.

Explanation:

When the incident ray enters the refracting surface, the ray bends from its original path because of the refractive index of the surface.

The angle of refraction is the angle which is made by the refracted ray and the perpendicular on the surface.

The perpendicular on the surface is known as Normal.

Hence, the angle of refraction is the angle between the refracted ray and the Normal.

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Two horses on opposite sides of a narrow stream are pulling a barge up the stream. Each hors pulls with a force of 720N. The rop
Sonja [21]
For the answer to the question above, each horse's force forms a right angle triangle with the barge and subtends an angle of 60/2 = 30°. The resultant in the direction of the barge's motion is:
Fx = Fcos(∅)
We can multiply this by 2 to find the resultant of both horses.
Fx = 2Fcos(∅)
Fx = 2 x 720cos(30)
Fx = 1247 N
8 0
3 years ago
at a particular instant a hot air balloon is 100m in the air and decending at a constant speed of 2m/s at this exact instant a g
labwork [276]
<h2>She will find the ball at a horizontal distance of 86.4 m from landed location</h2>

Explanation:

Consider the vertical motion of ball

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 2 m/s

        Acceleration, a = 9.81 m/s²  

        Displacement, s = 100 m      

     Substituting

                      s = ut + 0.5 at²

                      100 = 2 x t + 0.5 x 9.81 xt²

                      4.905t²  + 2t - 100 = 0

                     t = 4.32 s    or   t = -4.72 s

                    After 4.32 seconds the ball reaches ground.

Now we need to find horizontal distance traveled by ball in 4.32 seconds.

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 20 m/s

        Acceleration, a = 0 m/s²  

        Time, t = 4.32 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 20 x 4.32 + 0.5 x 0 x 4.32²

                      s = 86.4 m

She will find the ball at a horizontal distance of 86.4 m from landed location

5 0
4 years ago
Key Stage 3 Science - Physics
stira [4]

Answer:

F = 3750 N

Explanation:

Given that,

Pressure, P = 150 Pa

Area, a = 25m²

We need to find the force applied. We know that, pressure is equal to the force acting per unit area. It can be given by :

P=\dfrac{F}{A}\\\\F=P\times A\\\\F=150\ Pa\times 25\ m^2\\\\F=3750\ N

So, the required force is 3750 N.

5 0
3 years ago
How does an inclined plane affect the effort needed to move a load vertically?
lys-0071 [83]

Answer:

Decreases

Explanation:

"Effort" usually refers to the applied force.  An inclined plane decreases the force required while increasing the distance that the force is exerted over.  So even though there's less force needed, the amount of work stays the same.

4 0
3 years ago
In a physics laboratory experiment, a coil with 250 turns enclosing an area of 10.5 cm2 is rotated during the time interval 3.10
olchik [2.2K]
I guess the problem is asking for the induced emf in the coil.

Faraday-Neumann-Lenz states that the induced emf in a coil is given by:
\epsilon = -N \frac{\Delta \Phi}{\Delta t}
where
N is the number of turns in the coil
\Delta \Phi is the variation of magnetic flux through the coil
\Delta t is the time interval

The coil is initially perpendicular to the Earth's magnetic field, so the initial flux through it is given by the product between the magnetic field strength and the area of the coil:
\Phi_i = BA=(5.30 \cdot 10^{-5}T)(10.5 \cdot 10^{-4} m^2)=5.57 \cdot 10^{-8} Wb
At the end of the time interval, the coil is parallel to the field, so the final flux is zero:
\Phi_f = 0

Therefore, we can calculate now the induced emf by using the first formula:
\epsilon = -N  \frac{\Delta \Phi}{\Delta t}=- (250)  \frac{5.57 \cdot 10^{-8} Wb - 0}{3.10 \cdot 10^{-2} s} = -4.5 \cdot 10^{-4} V
7 0
4 years ago
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