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sineoko [7]
3 years ago
15

The photeselestric effect is observed when light of a sufficiently high frequency is focused onto a polished metal surface, emit

ting photo electrons with a kinetic energy that is given by the difference betweent he photon enerzy and the work function of the metal Historically, this effect was very confusing. For most wave phenon tand ight was wel known to behave as a wave phenomenon·ore nerease the increasing the amplitude of the wave. in the case of light, this means increase- there are simple more electrons emitted increasing the ententisty. However, with increased intensity, the kinetic energy of the photoelections emited does not Albert Einstein (1879-1955) offered a solution to this perplecing problem in hi 190s paper!,-Ober einen die Erzeugung und Verwandlung des Lichtes betrefflenden heuristischen Gesichtspunkr. Einstein proposed the following relationship: kinetic- hw-E work function energy of an emitted photoelectron, hw is the energy of a photon of frequency v and Ework function is the wrok function for the metal used in the ion is the energy required to remove an electron from the surface of the metal experiment. The work functi References Einstein, Albert (1905). The work function of cesium is 2.1 ev. What is this energy in? Hint: 1 ev-1,602 x 10-19) Annalen der Physik 17(6 132-148 (1905) 34x 1019) 21x 1019 64x 1019 43x 1019 19, 0 19, 0
Physics
1 answer:
Helga [31]3 years ago
8 0

Answer:

3.4\cdot 10^{-19} J

Explanation:

In order to convert the work function of cesium from electronvolts to Joules, we must use the following conversion factor:

1 eV = 1.6 \cdot 10^{-19} J

In our problem, the work function of cesium is

E=2.1 eV

so, we can convert it into Joules by using the following proportion:

1 eV : 1.6\cdot 10^{-19} J = 2.1 eV : x\\x=\frac{(1.6\cdot 10^{-19} J)(2.1 eV)}{1 eV}=3.4\cdot 10^{-19} J

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A 2.0 x 10^3-kilogram car travels at a constant speed of 12 meters per second around a circular curve of radius 30. meters. What
Vikentia [17]

The magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

<h3>Circular motion</h3>

From the question, we are to determine the magnitude of the centripetal acceleration.

Centripetal acceleration can be calculated by using the formula

a_{c} =\frac{v^{2} }{r}

Where a_{c} is the centripetal acceleration

v is the velocity

and r is the radius

From the given information

v = 12 \ m/s

and r = 30 \ m

Therefore,

a_{c} =\frac{12^{2} }{30}

a_{c} =\frac{144 }{30}

a_{c} = 4.8\ m/s^{2}

Hence, the magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

Learn more on circular motion here: brainly.com/question/20905151

4 0
2 years ago
A student has a mass (including clothes and shoes) of 65.0 kg. She drinks a 12 oz. can of soda, with a nutritional energy conten
Stella [2.4K]

She can climb 0.92 m without losing weight.

<u>Explanation</u>:

Gravitational potential energy is the energy consisting of the product of mass, gravity and height.

1 cal = 4184 J

140 cal = 585760 J

Energy = 585760 J,  m = 65.0 kg = 65000 g,   Efficiency = 20 %

                                 GPE = mgh

where m represents the mass

          g represents the gravity,

           h represents the height.

                             585760 = 65000 \times 9.8 \times h

                                        h = 0.92 m.

7 0
3 years ago
Which of the following electromagnetic waves has the shortest wavelength?
nignag [31]
Of the list, Ultraviolet waves have the shortest wavelength
7 0
3 years ago
Read 2 more answers
Can you help me with this??
s2008m [1.1K]

Answer:

i want to say flip the coins but im not really sure sry

Explanation:

3 0
3 years ago
Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm
Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: <em>Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.</em>

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

\eta\cdot \dot W = \dot K

\dot W = \frac{\dot K}{\eta} (Eq. 1)

Where:

\eta - Efficiency of fan, dimensionless.

\dot W - Electric power supplied fan, measured in watts.

\dot K - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2} (Eq. 2)

Where:

\rho_{a} - Density of air, measured in kilograms per cubic meter.

\dot V - Volume flow, measured in cubic meters per second.

A_{s} - Cross-sectional area of fan, measured in square meters.

If we know that \rho_{a} = 1.20\,\frac{kg}{m^{3}}, \dot V = 0.05\,\frac{m^{3}}{s}, \eta = 0.3 and A_{s} = 20\times 10^{-4}\,m^{2}, the power needed to be supplied by the fan is:

\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}

\dot K = 62.5\,W

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

5 0
3 years ago
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