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sergij07 [2.7K]
3 years ago
7

As light shines from air to water, the index of refraction is 1.02 and the angle of incidence is 38.0 °. What is the light's an

gle of refraction?
39.8°
37.1°
29.8°
32.5°
Physics
1 answer:
muminat3 years ago
5 0

Answer:

Light's angle of refraction = 37.1° (Approx.)

Explanation:

Given:

Index of refraction = 1.02

Base of refraction = 1

Angle of incidence = 38°

Find:

Light's angle of refraction

Computation:

Using Snell's law;

Sin[Angle of incidence] / Sin[Light's angle of refraction] = Index of refraction / Base of refraction

Sin38 / Light's angle of refraction = 1.02 / 1

Sin[Light's angle of refraction] = Sin 38 / 1.02

Sin[Light's angle of refraction] = [0.6156] / 1.02

Sin[Light's angle of refraction] = 0.6035

Light's angle of refraction = 37.1° (Approx.)

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An internal resistance of 5 ohm and a battery of 15 ohm is connected to a resistance of 20 ohm calculate the electric current
Allushta [10]

Answer:

.6 A

Explanation:

Battery 15 VOLTS

V = IR

V / R = I

15 / ( 5+20) = .6 amps

5 0
2 years ago
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

Download pdf
7 0
3 years ago
The rest deltoid row is a back exercise true or false
bixtya [17]
False because your deltoids are in your shoulders not your back
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3 years ago
A stationary siren emits sound of frequency 1000 Hz and wavelength 0.343 m. An observer who is moving toward the siren will meas
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Answer:

f>1000Hz and wavelength=0.343 m

Explanation:

We are given that

Frequency of stationary siren,f=1000 Hz

Wavelength of stationary sound,\lambda=0.343 m

When a observer is moving towards the siren then the frequency increases.

Therefore,an observer who is moving towards the siren measure a frequency >1000 Hz.

The wavelength depends upon the speed of source.

But we are given that siren is stationary.

Therefore, source is not moving and then the wavelength remains same.

f>1000Hz and wavelength=0.343 m

4 0
3 years ago
The greater the mass of an object, the greater the _____. (fill in question) ​
g100num [7]

Answer:

momentum

Explanation:

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