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Bingel [31]
3 years ago
9

A scuba diver uses his waterproof flashlight to shine a beam of light so that it strikes the surface of the water at an angle of

incidence θi. use snell's law to find the angle of incidence that would give an angle of refraction for the refracted ray to be directed right along the surface, and show that θi is the same as the critical angle for total internal reflection.
Physics
1 answer:
Mrrafil [7]3 years ago
8 0
Snell's law: n1Sinα=n2Sin β where α=Incidence angle, β=angle of refraction, n1 and n2 are the indices of refraction for water and air respectively.
Therefore,
Sinα=n2/n1 Sinβ For refracted ray to be along the surface of water, β=90° and thus Sinβ = 1

Sinα=n2/n1= 1/1.33 = 0.7519 => α=sin^-1 (0.7519) = 48.75°
When light moves from a medium of higher index of refraction (such as water) to medium of lesser index of refraction (such as air), the refracted ray is bend such that α is bigger than β. This  is internal refraction. At some value of α, β approaches 90°. This incidence angle is called critical incidence angle. Therefore, the current scenario is shows critical angle of incidence.
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A ball is being thrown straight upward and rises to a maximum height of 12.0m above its launch point. At what height above it’s
AlexFokin [52]

Answer:

9.0 m

Explanation:

Let the initial velocity be 'u'.

Given:

Final velocity is half of initial velocity.

Maximum height reached by ball (H) = 12.0 m

Acceleration of the ball is due to gravity (g) = -9.8 m/s²(Downward)

Now, first, we will find initial velocity of the ball using equation of motion given as:

v^2=u^2+2a(\Delta y)

For maximum height, final velocity is 0 as the ball stops at the maximum height temporarily. So, v=0\ m/s

Also, \Delta y=H=12\ m

Now, plug in all the values and solve for 'u'.

0^2=u^2+2(-9.8)(12)\\\\u^2=235.2\\\\u=\sqrt {235.2} =15.34\ m/s

Now, consider the motion of the ball till the velocity reaches half of initial velocity.

So, final velocity (v) = \frac{u}{2}=\frac{15.34}{2}=7.67\ m/s

Now, again using the same equation and finding the new height now. Let the new height be 'h'.

So, equation of motion is given as:

v^2=u^2+2ah\\7.67^2=15.34^2+2\times -9.8\times h\\58.83=235.2-19.6h\\\\19.6h=235.2-58.83\\\\19.6h=176.37\\\\h=\frac{176.37}{19.6}\approx9.0\ m

Therefore, the height reached by the ball when velocity is decreased to one-half of the initial velocity is 9.0 m.

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3 years ago
The diagram shows changes of state between solid, liquid, and gas. The atoms of a substance gain energy during a change of state
dolphi86 [110]
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3 years ago
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What is the potential energy of a 2,000-kg car parked at the top of a 30-m hill?
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Potential energy is a relative measure, so the answer is dependent on the assumptions we make. The potential energy in the car is going to be gravitational potential energy(PE). PE = mgh, where m is the mass, g is 9.8 m/s^2, and h is the height. So PE = 2000*9.8*h = 19600h. The final answer obviously depends on h. Most likely the problem is assuming that 30 meters under the top of the hill is considered 0 meters. Then h would be 30m and PE would equal 588 kJ.
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4 years ago
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A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t
tresset_1 [31]

Answer:

Part a)

Reading = 2.00 kg

Part b)

Reading = 2.00 kg

Part c)

Reading = 4.04 kg

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part b)

When elevator is decending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

F_{net} = ma

T - mg = ma

T = mg + ma

Reading is given as

Reading = \frac{mg + ma}{g}

Reading = 2.00\frac{9.81 + 10}{9.81}

Reading = 4.04 kg

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

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3 years ago
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