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Ostrovityanka [42]
3 years ago
10

The refractive index of water is greater than that of air but is less than that of glass or a diamond. Under which condition can

there be total internal reflection in water?
Physics
1 answer:
Xelga [282]3 years ago
5 0
For total internal reflection to happen, light must travel from a medium object of higher density to a medium of lower density. To make sure that total internal reflection will happen in water, then the light must originate from glass or diamond before entering water.
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According to the exercise principle of balance, a workout should __________.
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mina [271]

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Explanation:

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Thermal energy transfers from a cup of tea at 350 K to the hand holding it. <br> A)True<br> B)False
Alexandra [31]

Answer:

true

Explanation:

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3 years ago
Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
What is the work done when a 400.n force is use to lift a 400.n object 3.5 meters strait up
alexandr1967 [171]

Answer:

i think it's 1400

Explanation:

w = Force x displacement

w = 400 x 3.5

3 0
3 years ago
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