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o-na [289]
3 years ago
11

Light of a single frequency falls on a photo-electric material but no electrons are emitted . Electrons may be emitted if the __

___________

Physics
1 answer:
lesya692 [45]3 years ago
8 0
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
E=hf
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:
hf= \phi + K
where \phi 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.
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Can someone explain how the weight of the block is 10.26N, with reference to an appropriate law of motion?
spayn [35]

this process is called parellelogram method of resolving vectors.

5 0
3 years ago
28. Identify whether the following objects are in
Marianna [84]

Answer:

a. A baseball after it has been  hit - not in free fall

b. A rock that is thrown in the  air - not in free fall

c. The moon - free-fall

d. A paper airplane - not in free fall

e. A bird flying - not in free fall

Explanation:

  1. The free-fall is defined as the falling of an object due to the action of gravity. The object is not experiencing any other force neglecting the air resistance.
  2. If an object is in free-fall, the direction of its motion is directed towards the center of the earth. It does not have a horizontal component of velocity.
  3. If the body is under free-fall, but a centripetal force acts on it where it is equal to the gravitational force at that point. The object will have two components of velocity along the tangential line, perpendicular to the radius of the orbit.

a. A baseball after it has been  hit - not in free fall according to point 1 & 2.

b. A rock that is thrown in the  air - not in free fall according to point 1.

c. The moon - free-fall according to point 3.

d. A paper airplane - not in free fall according to point 1 & 2.

e. A bird flying - not in free fall according to point 1 & 2.

7 0
3 years ago
The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the
taurus [48]

The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the object sliding at a constant velocity once it starts.

The magnitudes of the required forces are different in these situations because the force of kinetic friction is less than the force of static friction. <em>(d)</em>

3 0
3 years ago
What is a main difference between Mars and Saturn?
tiny-mole [99]

Answer:

2

Explanation:

mars has a shorter year because when it revolves around the sun, it takes less time than, saturn who is millions of miles away from mars, longer. Also, mars is closer to the sun than saturn

8 0
3 years ago
Read 2 more answers
A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness
Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

\sigma_h = \frac{Pd}{2t}

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

\sigma_l = \frac{Pd}{4t}

The letters have the same meaning as before.

Then he hoop stress would be,

\sigma_h = \frac{Pd}{2t}

\sigma_h = \frac{75 \times 8}{2\times 0.25}

\sigma_h = 1200psi

And the longitudinal stress would be

\sigma_l = \frac{Pd}{4t}

\sigma_l = \frac{75\times 8}{4\times 0.25}

\sigma_l = 600Psi

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

\tau_{max} = \frac{\sigma_h}{2}

\tau_{max} = \frac{1200}{2}

\tau_{max} = 600Psi

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

6 0
3 years ago
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