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rosijanka [135]
3 years ago
13

An incident light ray strikes water at an angle of 20 degrees. The index of refraction of air is 1.0003, and the index of refrac

tion of water is 1.33. The angle of refraction rounded to the nearest whole number is °.
Physics
1 answer:
WINSTONCH [101]3 years ago
5 0
In order to find the angle of the refracted ray, we may use the Snell's law (also known as Snell–Descartes law and the law of refraction). This law states that the ratio of the sines of the angles of incidence and refraction is constant for a given wave when it passes through two different media such as water, glass, or air.
 In mathematical form, this is:

n₁sin(∅₁) = n₂sin(∅₂)
Where:n is the refractive index.

Plugging in the values given into the equation:
1.0003 * sin(20°) = 1.33 * sin(∅)
∅ = 14.91
The angle of refraction is 15°.
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A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.
Ray Of Light [21]

The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>
  • Mass of the box, m = 3.1 kg
  • Distance moved by the box, d = 2.5 m
  • Coefficient of friction, = 0.35
  • Inclination of the force, θ = 30⁰

<h3>What is work - done?</h3>
  • Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

W = (0) d \times cos(30)\\\\W = 0 \ J

The work done by the force of gravity is calculated as follows;

W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J

The work done on the box by the normal force is calculated as follows;

W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J

Learn more about work done here: brainly.com/question/8119756

8 0
2 years ago
Please help me this is timed . <br> Find x if a = 3.0 m/s^2
storchak [24]

Answer:

x = 50 N

Explanation:

Given that we have a net force, a mass, and acceleration, we can use the fundamental formula for force found in newton's second law which is F = m × a.

Given a mass of 150 kg, and an acceleration 3.0m/s². We can substitute these two values in our formula to calculate the magnitude of these forces or it's net force to identify the unknown force acting on our known force for this situation to work.

_______

F (Net force) = F2 (Second force which we are given) - F1 (First force) = m × a

m (mass which we are given) = 150 kg

a (acceleration which we are given) = 3.0m/s

________

So F = m × a → F2 - F1 = m × a →

500 - F1 = 150 × 3.0 → 500 - F1 = 450 →

-F1 = -50 → F1 = 50

5 0
3 years ago
Explain any two factors that affect the pressure at a point inside a liquid​
vredina [299]
The density of the body and the height or the depth of the body since the formula of liquid pressure is density x height gravity
4 0
2 years ago
Read 2 more answers
If
Aleksandr-060686 [28]

Answer:

A factor of 2*4 = 8

Explanation:

F_g = (G*m1*m2)/r^2  

where m1 and m2 are the two masses, G is Newton's gravitational constant, and r is the distance between the center of mass of the two objects.

So, if you double m1 and quadruple m2:

m1' = 2*m1

m2' = 4*m2

Then F_g' = (G*m1'*m2')/r^2 = (G*2*m1*4*m2)/r^2 = 8*(G*m1*m2)/r^2 = 8*F_g

3 0
2 years ago
Is acceleration occurring in this statement?
Burka [1]

Answer:

no because the car is in the rest so this not a acceleration

4 0
2 years ago
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