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Artemon [7]
3 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]3 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]3 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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A pilot heads his jet due east. The jet has a speed of 475 mi/h relative to the air. The wind is blowing due north with a speed
oksian1 [2.3K]

Explanation:

a. The velocity of the wind as a vector in component form will be represented as v vector:

    v=30j

b.The velocity of the jet relative to the air as a vector in component form will be represented as u vector

    u=475i

c. The true velocity of the jet as a vector will be represented as w:

  w=u+v

  w=475i+30j

d.  The true speed of the jet will be calculated as:

    IwI=\sqrt{(475)^2+(30)^2}

    IwI=\sqrt{225625+900}

    IwI=\sqrt{226525}

    IwI=476 mi/h

e. The direction of the jet will be:

tita=tan^{-1}\frac{30}{475}

tita=tan^{-1}(0.0632)

tita=3.62degrees,or,N86.38degreesS

7 0
3 years ago
2. A ball is dropped from rest. The acceleration due to gravity is 10m/s? and the time it
d1i1m1o1n [39]

Answer:

STEP BY STEP

V = S/T

V= 10/5

V = 2 m/s

3 0
3 years ago
Why does an object above the ground have P.E.?
Advocard [28]
This doesn't make sense
6 0
3 years ago
Read 2 more answers
A mass m is attached to an ideal massless spring. When this system is set in motion with amplitude a, it has a period t. What is
Luden [163]

The period will be the same if the amplitude of the motion is increased to 2a

What is an Amplitude?

Amplitude refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium.

Here,

mass m is attached to the spring.

mass attached = m

time period = t

We know that,

The time period for the spring is calculated with the equation:

T = 2\pi \sqrt{\frac{m}{k} }

Where k is the spring constant

Now if the amplitude is doubled, it means that the distance from the equilibrium position to the displacement is doubled.

From the equation, we can say,

Time period of the spring is independent of the amplitude.

Hence,

Increasing the amplitude does not affect the period of the mass and spring system.

Learn more about time period here:

<u>brainly.com/question/13834772</u>

#SPJ4

7 0
2 years ago
Show that the optimal launch angle for a projectile subject to gravity is 45o by carrying out the following steps: 6. Write down
Lorico [155]

Answer:

sin  2θ = 1    θ=45

Explanation:

They ask us to prove that the optimal launch angle is 45º, for this by reviewing the parabolic launch equations we have the scope equation

            R = Vo² sin 2θ / g

Where R is the horizontal range, Vo is the initial velocity, g the acceleration of gravity and θ the launch angle. From this equation we see that the sine function is maximum 2θ = 90 since sin 90 = 1 which implies that θ = 45º; This proves that this is the optimum angle to have the maximum range.

We calculate the distance traveled for different angle

          R = vo² Sin (2 15) /9.8

          R = Vo² 0.051 m

In the table are all values ​​in two ways

Angle (θ)                  distance R (x)

 0                 0                     0

15                 0.051 Vo²        0.5 Vo²/g

30                0.088 vo²        0.866   Vo²/g

45                0.102 Vo²        1   Vo²/g

60                0.088 Vo²      0.866   Vo²/g

75                0.051 vo²        0.5   Vo²/g

90                0                     0

See graphic ( R Vs θ)  in the attached ¡, it can be done with any program, for example EXCEL

6 0
3 years ago
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