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2 years ago

To understand the photoelectric effect physiciests should view light as having what type of properties?

Physics

1 answer:

fenix001 [56]2 years ago

5 0

Answer:

Particle

Explanation:

To understand photoelectric effect the physicists has to view light as a particle and not wave. The wave nature of the light could not explain the phenomenon of photoelectric effect.

Photoelectric effect is emission of electrons from a material when electromagnetic radiation(Light) is incident on it. These emitted electrons are called photoelectrons.

The explain this effect Einstein proposed that beam of light is can not be a wave propagating through space, but a swarm of discrete energy packets, called as photons.

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g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located

Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is: P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P /RT ⋯ ⋯( I )

Here

m is the mass of the air,

v is the volume of the air,

P is the atmospheric pressure,

T is the standard temperature at the atmospheric pressure and

R is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101 kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here A is the area of the windmill

Expression to calculate the A is

A = π /4 d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94 k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94 k g / s ( (6m/s)²/2 )

P w = 160144.92 W × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0

3 years ago

efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

PolarNik [594]

Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

$A_i=0.5m^2$ is the inlet area

$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy $h_o=1.0645\frac{KJ}{KgK}$

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

$\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}$

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

If we solve from the velocity v we have this:

$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

For the inlet case:

$v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}$

For the oulet case:

$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

7 0

3 years ago

When dogs see police dogs do they think "oh no its the cops"?

ira [324]

No they say "Watch out it's the fuzz"

4 0

2 years ago

Read 2 more answers

plz tell fast oktwo paragraphs describing the advantages and disadvantages of living in the town.plz tell

mars1129 [50]

Answer:

There are advantages and disadvantages to living in a small town. You are likely to have more frequent contact with neighbors in a small town than you would in a larger town or city, so it is important to take into account the pros and cons of having privacy. Also take into consideration other factors, including your plans for a family, the career opportunities, your finances, the schools and entertainment options. Your personality will determine whether a small town offers advantages or disadvantages.

Privacy

Many people who reside in small towns enjoy the closeness of knowing their neighbors and the details of neighbors' lives. If a baby is born, a child is ill or a relative dies, people in small towns often want to know about it and be there for you. This can be an advantage, especially if you are lacking a family of your own or enjoy a great deal of socializing. However, if you would rather keep the details of your life to yourself and turn your nose up at idle gossip, small town living might not be for you.

Finances

Jobs can be hard to come by in small towns. You might be required to commute out of town to work but then return to a peaceful environment at the end of a busy day. Housing might be cheaper in small towns, although commuting costs might be higher. Food and child care expenses might be reduced as people are more likely to have gardens and share produce or take turns watching each other's children.

Recreation

One of the disadvantages of a small town is not being close to amusement parks, zoos or museums. However, people in small towns tend to come together during town picnics, festivals or fairs. Depending on your preferences, small towns might provide a relaxed, slower pace when it comes to entertainment compared to the hustle and bustle of a city, where there is constant movement, commotion and entertainment at every corner.

Schools

In a small town, class sizes are usually smaller and teachers and students can get to know each other and their fellow classmates on a deeper level. Many small towns have schools that receive national academic awards or are the highest achieving district in their states. CNN Money ranked Louisville, Colorado as the number one small town in America in 2011 and the school system is academically ranked among the top three in the Denver area. However, some of these towns noted high taxes as a trade-off for the outstanding school systems.

Explanation:

3 0

2 years ago

In the united states a standard letter sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for

Andre45 [30]

This question is incomplete

Complete Question

In the United States, a standard letter-sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for a letter-sized piece of paper is different. The international standard is based on SI units: 21.0 cm wide by 29.7 cm long.

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

Answer:

a) 8.267721 inches ≈ 8.3 inches

b) 11.6929197 inches ≈ 11.7 inches

c) It's dimensions in inches for the international standard for letter - sized for paper = 8.3 wide inches by 11.7 inches long

d) The International standard letter - sized paper is longer.

Explanation:

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

21 cm =

Cross Multiply

21 cm × 0.393701inch/ 1 cm

= 8.267721 inches

Approximately 8.3 inches

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

29.7 cm =

Cross Multiply

29.7 × 0.393701 inch/ 1 cm

= 11.6929197 inches

Approximately 11.7 inches

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

The international standard for a letter-sized has dimensions 21.0 cm wide by 29.7 cm long.

Where

21.0cm = 8.267721 inches

≈ 8.3 inches

29.7cm = 11.6929197 inches

≈ 11.7 inches

Hence, it's dimensions in inches = 8.3 inches by 11.7 inches.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

U.S letter - sized paper = 8.5 inches wide by 11 inches long

International standard letter- sized paper = 8.3 wide inches by 11.7 inches long.

Hence, the International standard letter - sized paper is longer.

5 0

3 years ago