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nexus9112 [7]
3 years ago
14

NEED HELP ASAP! DUE NOW!!

Physics
1 answer:
docker41 [41]3 years ago
4 0
Oh okay the first person who answered is incorrect but


0—> PE=0J

1—>PE=1J

20—>PE=20J

50—>PE=50J

100—>PE=100J
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The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
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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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3 years ago
A child wearing socks is trying to see how far he can slide across his kitchen floor. If
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Answer:

The coefficient of static friction between the ground and the soles of a runner’s shoes is 0.98. What is the maximum speed in which the runner can accelerate without slipping if they have a mass of 73 kg?

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Is Smoke Abiotic or Biotic?
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Answer: Smoke is Abiotic because it is not a living thing.

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What are the unit for acceleration
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In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr
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Answer:

a) v_p=9.35m/s

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From the question we are told that:

Open cart of mass   M_o=50.0 kg

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Distance of chute from bottom of cart   d_x=4.00m

a)

Generally the equation for work energy theory is mathematically given by

  \frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2

Therefore

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