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Paraphin [41]
4 years ago
14

A tank holds a 2.38-m thick layer of oil that floats on a 1.24-m thick layer of brine. Both liquids are clear and do not intermi

x. Point O is at the bottom of the tank, on a vertical axis. The indices of refraction of the oil and the brine are 1.27 and 1.81, respectively. A ray originating at O reaches the brine-oil interface at the critical angle. What is the distance of this point from the axis?
Physics
1 answer:
kozerog [31]4 years ago
6 0

Answer:

1.22m

Explanation:

Since

sinθ  =  refraction-of-the-oil/refraction-of-the-brine =1.27/1.81 = 0.702

θ = sin^-^1(0.702)

Hence

Critical angle = θ = 44.58°

tan(θ) = d/1.24

tan(44.58°) = d/1.24

Hence, 0.98 = d/1.24

The distance d = 0.98 x 1.24 = 1.22m

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Answer:

V_2 = 0.125 m^3

Work done =  = 5 kJ

Explanation:

Given data:

volume of nitrogen v_1 = 0.08 m^3

P_1 = 150 kPa

T_1 = 200 degree celcius = 473 Kelvin

P_2 = 80 kPa

Polytropic exponent n = 1.4

\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}

putting all value

\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}

\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS

polytropic process is given as

P_1 V_1^n = P_2 V_2^n

150\times 0.08^{1.4} = 80 \times V_2^{1.4}

V_2 = 0.125 m^3

work done = \frac{P_1 V_1 -P_2 V_2}{n-1}

= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}

                  = 5 kJ

4 0
3 years ago
A sealed container filled with gas is heated. What happens?
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Answer:

when the ground is very hot and the air is cool.

Explanation:

The hot earth warms a layer of air right above the ground. Light is refracted as it passes through the cool air and onto the hot air sheet (bent). A coating of very warm air near the earth bends the light from the sky almost into a U-shaped bend.

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An object of mass 0.400.40 kg, hanging from a spring with a spring constant of 8.08.0 N/m, is set into an up-and-down simple har
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Answer:

Acceleration will be equal to 2m/sec^2      

Explanation:

We have given mass of the object m = 0.4 kg

Spring constant k = 8 N/m

Maximum displacement of the spring is given x = 0.1 m

From newton's law force is equal to F=ma.....eqn 1

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