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Amanda [17]
3 years ago
9

The Doppler effect is an effect produced by a moving source of sound or electromagnetic waves due to the relative motion of a so

urce or a receiver. If an emergency vehicle approaches Bob and moves away from Jill, how does the frequency of the siren change? A) Neither Bob or Jill perceive any change in the frequency of the siren. B) As an emergency vehicle approaches Bob the frequency appears to increase. C) As an emergency vehicle moves away from Jill the frequency appears to increase. D) Bob does not notice any change until the vehicle passes him and the siren increases.
Physics
2 answers:
Len [333]3 years ago
8 0

Answer:

B) As an emergency vehicle approaches Bob the frequency appears to increase.

Explanation:

As mentioned Doppler effect is related to the waves produced by a moving source. When moving towards the observer, the frequency of the wave will seem to increase and while moving away from the observer, the frequency will seem to decrease.

Here, the vehicle is moving towards Bob and away from Jill. This means, the frequency of the siren will seem to increase while approaching Bob and frequency will seem to decrease while moving away from Jill.

xxMikexx [17]3 years ago
4 0
B. The waves would shorten as the vehicle got closer
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The components of a 10.8-meters-per-second velocity at an angle of 34.° above the horizontal are
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Answer:

6.0 m/s vertical and 9.0 m/s horizontal

Explanation:

For the vertical component, we use the formula:

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And for the horizontal component, we use the formula:

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3 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:

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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

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Answer: b. Throw it directly away from the space station.

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