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Artemon [7]
4 years ago
8

In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the met

al are held constant. Assuming that the light incident on the metal surface causes electrons to be ejected from the metal, what happens if the intensity of the incident light is increased?
Physics
2 answers:
Novosadov [1.4K]4 years ago
7 0

Answer:

Explanation:

When a light beam of suitable frequency falls on metal surface, the electrons from the metal surface ejects, this phenomenon is called photo electric effect and the particular minimum frequency is called cut off frequency or threshold frequency.

When the intensity of incident radiation increases, the photo electric current also increases.  

mixer [17]4 years ago
5 0

Answer:

The kinetic energy of the ejected electrons increases.

Explanation:

As we know that electrons are only ejected from a metal surface if the frequency of the incident light increases the work function of the metal. If the frequency of the incident light is less than the work function of the metal no matter how intense the beam the electrons will not be ejected from the surface.

Using conservation of energy principle we have

E_{incident}=h\nu +\frac{1}{2}mv^{2}

If we increase the intensity  of incident light the term on the LHS of the above equation increases this increase appears in the kinetic energy term in RHS of the equation since h\times \nu remains constant.

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You drive your car at a speed 40km/h for 1 hour, then slow down to 30km/h for the next 20km. How far did you drive, and what was
Bingel [31]
Speed=distance/time, rearranging gives speed*time=distance therefore the first section of the journey yielded a total distance of 40km, the second journey was stated to be 20km which was completed in (20/30)*60=40minutes. Therefore the total journey was 60km which was completed in 1hr40 minutes so the average speed was (60/1.66..)=36km/h.
8 0
3 years ago
Which of the following has greater number of hydrogen molecule ? 9 gm of CH4 or 10gm of NH3​
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Explanation:

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Now set the tension to low and wiggle the wrench to create more waves. Can you explain how moving the first point on the string,
Hatshy [7]

Answer:

When the string moves, it creates a very small change in the distance to the next point, th

Explanation:

When the string moves, it creates a very small change in the distance to the next point, this generates a restoring force that tends to push the string back, this small disturbance propagates along the string and is what creates the pulse.

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4 0
3 years ago
Read 2 more answers
A 1.5 kg bird is gliding at a height of 12 m with a speed of 3.8m/s. What is the kinetic energy of the bird, to the nearest joul
Georgia [21]

A 1.5 kg bird is gliding at a height of 12 m with a speed of 3.8m/s. The kinetic energy of the bird is 10.83 joules.

Explanation:

Kinetic energy can be defined as,The kinetic energy (KE) of an object is the energy that the object possesses due to its motion.

The Kinetic energy can be calculated by using formula,

Kinetic Energy: KE = 1/2 (mv 2)

Where, m = Mass, v = Velocity.

Here in this case the bird mass is 1.5kg and is gliding with velocity 3.8m/s

hence, KE= 1/2*(1.5)×(3.8)^2

                   =0.5×1.5×3.8×3.8

                   =10.83Joules

                   

                   

5 0
3 years ago
A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel
Dafna1 [17]

Answer:

1.69\cdot 10^{10}J

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

E=-G\frac{mM}{2r}

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

M=5.98\cdot 10^{24}kg is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m

So the initial total energy is

E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J

When the satellite hits the ground, it is now on Earth's surface, so

r=R=6370 km=6.37\cdot 10^6 m

so its gravitational potential energy is

U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J

And since it hits the ground with speed

v=1.90 km/s = 1900 m/s

it also has kinetic energy:

K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J

So the total energy when the satellite hits the ground is

E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J

8 0
3 years ago
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