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Rainbow [258]
2 years ago
14

Experiment to determine the refractive index of a glass prism FOCAL LENGTH = 9.5cm i° e° d° 30 43 69 40 41 61 45 39 56 50 37 48

65 41 37 plot a graph of d° against i° from the graph Determine; (i) the minimum deviation and the corresponding angle of incidence (ii) the maximum deviation and the corresponding angle of incidence n=sin⁡(A+D÷2)÷(sinA÷2) Determine the error in n and explain why it is not advisable to use small values of i° in performing this experiment. THANK YOU SO MUCH
Physics
1 answer:
yarga [219]2 years ago
3 0
We are given the data for the angle of incidence and angle of refraction.
The first part of the problem is to plot the deviation and the angle of incidence. The minimum deviation is 37 with 65 degrees as the corresponding angle of incidence. The maximum deviation is 69 with a corresponding angle of incidence of 30 degrees. It is not advisable to use small values for the angle of incidence since it would result to a higher deviation from Snell's Law.
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A ball is thrown straight up. What are the velocity and acceleration of the ball at the highest point in its path?
zubka84 [21]

Answer:

b. v = 0, a = 9.8 m/s² down.

Explanation:

Hi there!

The acceleration of gravity is always directed to the ground (down) and, near the surface of the earth, has a constant value of 9.8 m/s². Since the answer "b" is the only option with an acceleration of 9.8 m/s² directed downwards, that would solve the exercise. But why is the velocity zero at the highest point?

Let´s take a look at the height function:

h(t) = h0 + v0 · t + 1/2 g · t²

Where

h0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity

Notice that the function is a negative parabola if we consider downward as negative (in that case "g" would be negative). Then, the function has a maximum (the highest point) at the vertex of the parabola. At the maximum point, the slope of the tangent line to the function is zero, because the tangent line is horizontal at a maximum point. The slope of the tangent line to the function is the rate of change of height with respect to time, i.e, the velocity. Then, the velocity is zero at the maximum height.

Another way to see it (without calculus):

When the ball is going up, the velocity vector points up and the velocity is positive. After reaching the maximum height, the velocity vector points down and is negative (the ball starts to fall). At the maximum height, the velocity vector changed its direction from positive to negative, then at that point, the velocity vector has to be zero.

8 0
3 years ago
How can you verify the archimedes principle?​
Ipatiy [6.2K]

Answer:

It is found that W1 - W2 loss in weight of solid when immersed in water is equal to the weight of the water displaced by the body. This verifies Archimedes' principle.

6 0
2 years ago
If you were trying to build a soundproof room, which of the following materials would you choose in order to absorb the most sou
leonid [27]

Explanation :  

Absorption coefficient of a material determines how much sound is absorbed by the material.

To build a soundproof room, Heavy curtains and carpet can be used. They reduce reverberation.

Reverberation means an echoing sound which persists for some time. For example, when we bang on a huge piece of metal, we hear the reverberation even after we stop banging.  

Hence, option (A) and (D) are correct.

5 0
3 years ago
Read 2 more answers
9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o
Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).

Now the force of friction is F_s=\mu*N, where N is the normal force and from the diagram it is F_y=mg*cos(\theta).

Thus F_s=\mu*N=\mu*mg*cos(\theta).

Therefore,

F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).

Substituting the value for F_H,m,\mu, and \:\theta we get:

F_{tot}= -38.63N.

Now acceleration is simply

a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.

The negative sign indicates that the acceleration is directed up the incline.

Part B:

d=\frac{1}{2} at^2

Which can be rearranged to solve for t:

t=\sqrt{\frac{2d}{a} }

Substitute the value of d=0.50m and a=7.7m/s and we get:

t=0.36s.

which is our answer.

Notice that in using the formula to calculate time we used the positive value of a, because for this formula absolute value is needed.

5 0
3 years ago
A model engine accelerated forward from rest along a straight track at 10.0 m/s2 for 3.0 seconds. It then accelerates coward at
Mrac [35]

Answer:

17.2 seconds

Explanation:

Given:

v₀ = 0 m/s

a₁ = 10.0 m/s²

t₁ = 3.0 s

a₂ = 16 m/s²

t₂ = 5.0 s

a₃ = -12 m/s²

v₃ = 0 m/s

Find: t

First, find v₁:

v₁ = a₁t₁ + v₀

v₁ = (10.0 m/s²) (3.0 s) + (0 m/s)

v₁ = 30 m/s

Next, find v₂:

v₂ = a₂t₂ + v₁

v₂ = (16 m/s²) (5.0 s) + (30 m/s)

v₂ = 110 m/s

Finally, find t₃:

v₃ = a₃t₃ + v₂

(0 m/s) = (-12 m/s²) t₃ + (110 m/s)

t₃ = 9.2 s

The total time is:

t = t₁ + t₂ + t₃

t = 3.0 s + 5.0 s + 9.2 s

t = 17.2 s

Round as needed.

4 0
3 years ago
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