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frozen [14]
3 years ago
9

Which visible colors of light has the highest frequency?

Physics
1 answer:
Valentin [98]3 years ago
7 0

Answer:

Violet color has the highest frequency.

Explanation:

Light: it can behave as particle and wave. It has both natures. Frequency is the characteristic of the wave. It is defined as how many waves can pass through at a point in a given interval of time. It is measured in hertz.

Red: It lies in the visible region of the light. it has the lowest frequency in the visible region of the light.

Violet:  It lies in the visible region of the light. Every wave has some energy. The energy of the wave is directly proportional to the frequency of the wave. More the energy of the wave then more will be the frequency of the wave. Violet color has more energy. It has the highest frequency.

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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
4 years ago
Two object, one of mass m and the other of mass 2m, are dropped from the top of a building. if there is no air resistance, when
ra1l [238]

Answer:

They hit the ground at the same time

Explanation:

Mass doesn't matter in free fall. As long as they were released with the same velocity and the same height, they'll hit the ground at the same time

3 0
3 years ago
The drawing shows three identical springs hanging from the ceiling. Nothing is attached to the first spring, whereas a 4.5 n blo
Naya [18.7K]

<Continuation of the question>

(a) the spring constant (in N/m) and

(b) the weight of the block hanging from the third spring.

the distance for the first spring is 20cm, the second 35cm, the third 50cm.

Answer:

a) To find the spring constant, we'll use the formula

F=kx, if we make k the subject we'll get

k=F/x, where F=4.5N, x = 35cm - 20cm = 15cm = 0.15m

k=F/x = 4.5N/0.15m =  30N/m is the Answer

b) to find the weight of block hanging on third spring

we use the formula F=kx

where k = 30N/m, x=50cm-20cm=30cm=0.30m

F=kx = (30N/m)*(0.30m) = 9N  is the Answer

4 0
3 years ago
While tuning a string to the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string.
lara31 [8.8K]

Answer:

a)the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b) 526hz

c)0.989 or a 1.14% decrease in tension

Explanation:

a) While tuning a string at 523 Hz,piano tuner hears 2.00 beats/s between a reference oscillator and the string.

The possible frequencies of the string can be calculated by

fl=f' - B

where

fl= lower limit of the possible frequency

f'= frequency of the string

B= beat heard by the tuner

fl= 523hz + Or - (2beats/secs * 1hz/1beat per sc)

fl= 521hz or 525hz

So the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b)fl=f' - B

523hz= f' - 3

f'= 523 + 3= 526hz

c) The tension is directly proportional to the square of the frequencies

T1/T2 =f1^2/f2^2

523^2 / 526^2 = 0.989 or a 1.14% decrease in tensio

6 0
3 years ago
The dwarf planet Pluto was discovered in 1930. Since that time, which jovian planet has completed a full revolution around the S
Eva8 [605]

Answer:

Explanation:

Uranus takes about 84 years to circle the sun.

Pluto's existance has been known for 92 years. Uranius has completed 1 complete revolution. The numbers are reasonably close together and so Uranius is the answer.

6 0
3 years ago
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