Light of a single frequency falls on a photo-electric material but no electrons are emitted . Electrons may be emitted if the __
___________
1 answer:
Correct answer is "frequency of the light is increased".
The light that hits the photo-electric material consists of several photons, each one carrying an energy of
where h is the Planck constant while f is the light frequency. Part of this energy is used to extract the photo-electrons from the material, and part it is converted into kinetic energy K of the electron:
where
is the work function of the material, which is the amount of energy needed to extract the photo-electron.
If no photo-electrons are emitted, it means that the energy given by the photon is not enough to extract the photo-electron. But we see that the energy of the photon is directly proportional to the frequency of the light f, so in order to increase the energy, we should increase the frequency of the light.
The light that hits the photo-electric material consists of several photons, each one carrying an energy of
where h is the Planck constant while f is the light frequency. Part of this energy is used to extract the photo-electrons from the material, and part it is converted into kinetic energy K of the electron:
where
If no photo-electrons are emitted, it means that the energy given by the photon is not enough to extract the photo-electron. But we see that the energy of the photon is directly proportional to the frequency of the light f, so in order to increase the energy, we should increase the frequency of the light.
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Answer:
The maximum mass the bar can support without yielding = 32408.26 kg
Explanation:
Yield stress of the material () = 200 M Pa
Diameter of the bar = 4.5 cm = 45 mm
We know that yield stress of the bar is given by the formula
Yield Stress =
⇒ =
---------------- (1)
⇒ Area of the bar (A) = ×
⇒ A = ×
⇒ A = 1589.625
Put all the values in equation (1) we get
⇒ = 200 × 1589.625
⇒ = 317925 N
In this bar the is equal to the weight of the bar.
⇒ =
× g
Where is the maximum mass the bar can support.
⇒ =
Put all the values in the above formula we get
⇒ =
⇒ = 32408.26 Kg
There fore the maximum mass the bar can support without yielding = 32408.26 kg
The hawk’s centripetal acceleration is 2.23 m/s²
The magnitude of the acceleration under new conditions is 2.316 m/s²
radius of the horizontal arc = 10.3 m
the initial constant speed = 4.8 m/s
we know that the centripetal acceleration is given by
=
= 23.04/10.3
= 2.23 m/s²
It continues to fly but now with some tangential acceleration
= 0.63 m/s²
therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration
so
=
=
= 2.316 m/s²
So the magnitude of net acceleration will become 2.316 m/s².
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Answer:
19.3m/s
Explanation:
Use third equation of motion
where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity
insert values to get answer