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postnew [5]
3 years ago
9

If the frequency of this beam is increased while the intensity is held constant, does the number of electrons ejected per second

from the metal surface increase, decrease, or stay the same?
Physics
1 answer:
Ivan3 years ago
8 0

Answer:

if the intensity of photons is constant then number of ejected electrons will remain same

Explanation:

As per photoelectric effect we know that when light of sufficient frequency fall on the surface of metal then electrons get ejected out of the surface with certain kinetic energy

Here the energy of photons is used to eject out the electrons from metal surface and to give the kinetic energy to the ejected electrons

so we have

h\nu = W + KE

here W = work function of metal which shows the energy required to eject out electrons from metal surface

KE = kinetic energy of ejected electrons

now if we increase the frequency of the photons that incident on the metal surface then in that case the incident energy will increase

So the electrons will eject out with more kinetic energy while if the number of photon is constant or the intensity of photons is constant then number of ejected electrons will remain same

You might be interested in
Calculate the potential difference of a wire of 10 ohm , through which 5 A of current is flowing .
HACTEHA [7]
Resistance= Potential Difference/Current
10 Ohm= PD/ 5A
PD= 10 Ohm × 5A
PD=50
Therefore, Potential Difference is B. 50 V
5 0
3 years ago
How much work is needed to stretch this spring a distance of 5.0 cm , starting with it unstretched?
olganol [36]

The spring's work output while it is not extended is 5 J.

<h3>What is The Hooke's law ?</h3>

According to Hooke's law, as long as the elastic limit is not exceeded, the extension of a particular material is directly proportional to the force applied. Hence, we must be aware of that F = Ke

F = Applied force

Force constant is K.

Extension = e

F = applied force

Force constant is K.

Extension = e

Using the graph

K = F/e

F = 200N

E equals 5 cm or 5 * 10-2 m

K = 200N/ 5 * 10^-2 m

K = 4000 N/m

Now;

Work = 1/2Ke^2

Work = 0.5 * 4000 N/m * (5 * 10^-2 m)^2

Work = 5 J

To know  more about Hooke's Law visit:

brainly.com/question/13348278

#SPJ4

4 0
1 year ago
Please HELP ME If the motion of an object changes, what must be true about the forces acting on that object?
atroni [7]

Answer:

If there is a net force acting on an object, the object will have an acceleration and the object's velocity will change. ... Newton's second law states that for a particular force, the acceleration of an object is proportional to the net force and inversely proportional to the mass of the object.

Explanation:

4 0
3 years ago
Read 2 more answers
Which of the following items uses a wheel and an axle?
MAVERICK [17]
The answer would be a bicycle as it uses both wheels and an axel
6 0
4 years ago
A 1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the
Paul [167]

Answer:

<em>a) 3344 N</em>

<em>b) 3344 N</em>

Explanation:

This is the complete question

1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected.  A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units.  B. What is the magnitude of the force of the truck on the car?

Mass of the car = 1100 kg

Mass of the truck = 2200 kg

Force exerted on the ground by the car = 5000 N

The total mass in the system = 1100 + 2200 = 3300 Kg

Total force in the system = 5000 N

Recall that the force in the system = mass x acceleration

therefore,

5000 = 3300 x a

Total acceleration in the system = 5000/3300 = 1.52 m/s^2

The force on the truck individually fro the car, will be the product of this acceleration and its mass

Force on the truck = 2200 x 1.52 = <em>3344 N</em>

b) Force on the car From the truck will be equal to this force but will act in the opposite direction.

Force on the car from the truck is <em>3344 N </em>

7 0
3 years ago
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