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Julli [10]
1 year ago
12

The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum

kinetic energy of the electrons when radiation of wavelength 198nm is used? c. It is observed that the electrons may be prohibited from reaching the anode by applying a stopping potential. What is the stopping potential in the case when radiation of wavelength 198nm is used?
Physics
1 answer:
Tanya [424]1 year ago
5 0

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

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liubo4ka [24]

Answer:

T = 7.64 kN

F_y = 0.52 kN(Downwards)

F_x = 3.23 kN (Towards Left)

Explanation:

As we know that beam is in equilibrium

So here we can use torque balance as well as force balance for the beam

Now by torque balance equation at the pivot we can say

F(4.50 cos\theta) + mg(2cos\theta) = T \times 3

As we know that

mg = 1.40 kN

F = 5 kN

so we will have

5 kN(4.50 cos25) + 1.40 kN(2 cos25) = 3 T

T = 7.64 kN

Now force balance in vertical direction

F + mg = Tsin65 + F_y

5 + 1.40 = 7.64 sin65 + F_y

F_y = 0.52 kN(Downwards)

Force balance in horizontal direction

F_x = T cos65

F_x = 7.64 cos65

F_x = 3.23 kN (Towards Left)

7 0
2 years ago
Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in scientific notation?
Oksi-84 [34.3K]

Answer:

2.32\times 10^{-11}

Explanation:

First number is 4.48\times 10^{-8}

Second number is 5.2\times 10^{-4}

We need to multiply the two numbers.

4.48\times 10^{-8}\times 5.2\times 10^{-4}=(4.48\times 5.2)\times 10^{(-8-4)}\\\\=23.296\times 10^{-12}

In scientific notation : 2.32\times 10^{-11}

Hence, this is the required solution.

8 0
2 years ago
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Since we are given the density and volume, then perhaps we can determine the amount in terms of the mass. All we have to do is find the volume in terms of cm³ so that it will cancel out with the cm³ in the density. The conversion is 1 ft = 30.48 cm. The solution is as follows:

V = (14 ft)(15 ft)(8 ft)(30.48 cm/1 ft)³ = 0.0593 cm³

The mass is equal to:
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7 0
2 years ago
4. While broadcasting a football game, the announcer exclaimed, "I can't believe it. Carl James just scored a touchdown. That's
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D. Carl doesnt score touchdowns very often.
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2 years ago
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(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind
Vilka [71]
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
F_f = \mu N
where \mu is the coefficient of friction, while N is the normal force. So we have:
F=\mu N
since we know that F=300 N and \mu=0.3, we can find N, the magnitude of the normal force:
N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is \mu=0.03 due to the presence of the oil. Therefore, we have:
N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N
8 0
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