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Maksim231197 [3]
2 years ago
12

Light with energy equal to three times the work function of a given metal causes the metal to eject photoelectrons. What is the

ratio of the maximum photoelectron kinetic energy to the work function?​
Physics
1 answer:
Mrac [35]2 years ago
5 0

The ratio of the maximum photoelectron kinetic energy to the work function will be 3:1.

<h3 /><h3>What is the photoelectric effect?</h3>

When a medium receives electromagnetic radiation, electrostatically charged particles are emitted from or inside it.

The emission of ions from a steel plate when light falls on it is a common definition of the effect. The substance could be a solid, liquid, or gas; and the released particles could be protons or electrons.

A particular metal emits photoelectrons when exposed to light with energy three times its work function:

\rm KE=3 \phi

The ratio of the maximum photoelectron kinetic energy to the work function will be;

R=\frac{E}{\phi} \\\\ R=\frac{3 \phi}{\phi} \\\\ R= 3

Hence, the ratio of the maximum photoelectron kinetic energy to the work function will be 3:1.

To learn more about the photoelectric effect refer to the link;

brainly.com/question/9260704

#SPJ1

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A boat is stationary at 12 meters away from a dock. The boat then begins to move toward the dock with
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Answer: 2.19 seconds

Explanation:

<u>Given:</u>

Initial speed, u = 0 m/s

Acceleration of the boat, a = 5 m/s^2

Distance between the boat and dock, d = 12 m

Using the third kinematics equation to solve for time:

d = u*t + (1/2)*a*t^2

12 = 0*t + (1/2)*5*t^2

t = sqrt (12*2/5)

t = 2.19 seconds

Therefore, it will take the boat approximately 2.19 seconds to reach the dock

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2 years ago
Which of these is NOT a reason why the geocentric model of the solar system was once commonly accepted as the correct model?
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3 years ago
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1. Ignore friction and determine an expression for the distance d the boxes travel before coming to rest. Choose the floor as th
Nadusha1986 [10]

Newton's second law allows us to find the results for the displacement and work on box 1 are:

       1) The displacement is   x = \frac{v_o^2 m_2}{2(m_1+m_2)} g

      2) The work is  W= \frac{1}{2} m_1v_o^2

1) Newton's second law says that force is directly proportional to the mass and acceleration of bodies.

         F = ma

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the bodies

The reference system is a coordinate system with respect to which the decompositions of the forces are carried out and all the measurements are made, in this case we take a system where the x axis is horizontal, the y axis is vertical and the zero of the system is located at soil

In the attached we can see a free body diagram of the system where the external forces are indicated, let's apply Newton's second law to body 1

body 1

x-axis

          -T = m₁ a

body 2

 y-axis

           T - W₂ = m₂ a

           W₂ = m₂ g

let's solve the system

        - m₂ g = (m₁ + m₂) a

           a = - \frac{m_2}{m_1+m_2} \ g- m2 / m1 + m2 g

Kinematics studies the movement of bodies, let's use the expression

            v² = v_o^2 - 2 a x

When the body stops the velocity is zero

            x = \frac{v_o^2}{2a}

We substitute

            x = \frac{v_o^2}{2} \frac{m_2}{m_1+m_2} \ g

Since the two boxes are connected by a rope, they both travel the same distance

           

2) They ask for Tension work on box 1

Work is defined by the scalar product of force and displacement

        W = F. d

Where the bold letters indicate vectors, W is the work that is a scalar, F the force and d the displacement

We can write this expression by developing the dot product

       W  = F d cos θ

Where θ is the angle between force and displacement.

In this case the force is the tension of the rope, from the attached graph we see that the force is directed to the right and the displacement is to the left, therefore the angle is 180º and the cos  180 is equal to -1

We look for the tension from Newton's second law for box 1

          T = - m₁ a

Let's substitute

          T =- m_1 \ ( - \frac{m_2}{m_1+m_2} ) \ g  

          T = \frac{m_1m_2}{m_1+m_2} \ g

         

We calculate the work

         W = - \frac{m_1m_2}{m_1+m_2 } \ g ( \frac{v_o^2 (m1+m_2)}{2 m_2 g} )

         W = - ½ m₁ v₀²

In conclusion using Newton's second law we can find the results for the displacement and work on box 1 are:

1)  The displacement is  x= \frac{v_o^2 m_2}{2(m_1+m_2)} g

2) The work is  W = - ½ m₁ v₀²

Learn more here: brainly.com/question/17290735

3 0
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An object is moving east with a constant speed of 30 m/s for 5 seconds. What is the object's acceleration
pentagon [3]

Answer:

<h3>The answer is 6 m/s²</h3>

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

a =  \frac{v}{t}  \\

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v is the velocity

t is the time

We have

a =  \frac{30}{5}  \\

We have the final answer as

<h3>6 m/s²</h3>

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Answer:

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It has been a long time since we have spent some time together. If you are free, I would welcome to have your company this weekend. Why don’t you come over to my house and spend a day or so with me?

I am anxiously waiting for your reply.

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