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notsponge [240]
2 years ago
8

BRAINLIST A wave travels at a constant speed. How does the wavelength change if the

Physics
1 answer:
Studentka2010 [4]2 years ago
3 0
I think it will be (c)
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cargo planes hold cargo so there hevier

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When force is applied to a breaker bar the torque can be calculated by multiplying the length of the lever by the?
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When a force applied to a breaker bar the torque can be calculated by multiplying the<u> length of the lever</u> by the tangential component of force on the lever.

<h3>What is torque?</h3>

Torque is the <u>rotating equivalent</u> of force in physics and mechanics. Depending on the subject of study, it is also known as the moment, moment of force, rotating force, or turning effect. It illustrates how a force can cause a change in the body's rotational motion.

Torque is given by the formula :

                          α = r x F ( bold letters represent vector quantities)

The S.I. unit for torque is :  N - m ( Newton - meter)

<h3>How do we define 1 N-m of torque?</h3>

The newton-metre is a torque unit (also known as a moment) in the SI system. The torque produced by a one newton force applied <u>perpendicularly to the end of a one metre long</u> moment arm is known as a newton-metre.

To learn more about torque:

brainly.com/question/14970645

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2 years ago
Which statement best explains the path light takes as it travels? A. Light takes a curved path through matter and takes a straig
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i show the thing. i hope this helps

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If your face is 62 cm in front of a plane mirror, where is the image of your face located?
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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
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