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aliya0001 [1]
3 years ago
9

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

metal 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?A. The work function of the metal decreases. B. The number of electrons emitted from the metal per second increases. C. The maximum speed of the emitted electrons increases. D. The stopping potential increases.2. In another experiment, the intensity of the incident light and the temperature of the metal are held constant. Assuming that the initial light incident on the metal surface causes electrons to be ejected from the metal, what happens if the frequency of the incident light is increased? A. The work function of the metal increases. B. The number of electrons emitted from the metal per second increases. C. The maximum speed of the emitted electrons increases. D. The stopping potential increases.
Physics
1 answer:
shutvik [7]3 years ago
4 0

Answer: 1. B. The number of electrons emitted from the metal per second increases.

2. The maximum speed of the emitted electrons increases.

The stopping potential increases

Explanation:

Photoelectric effect is simply referred to as the emission of electrons that occurs when there's an electromagnetic radiation. An example of such electromagnetic radiation is when material is being hit by light.

Assuming that the light incident on the metal surface causes electrons to be ejected from the metal, the number of electrons emitted from the metal per second increases if the intensity of the incident light is increased.

Also, if the initial light incident on the metal surface causes electrons to be ejected from the metal, the maximum speed of the emitted electrons increases and the stopping potential increases.

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Answer:

 v =7.1 m/s

Explanation:

Given that

u = 3.35 m/s

t= 5 s

a= 0.75 m/s²

The final velocity = v

We know v = u +at

v=final velocity

u=initial velocity

Now by putting the values in the above equation

v = 3.35 + 0.75 x 5 m/s

v =7.1 m/s

Therefore the final velocity will be 7.1 m/s

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The hardest known natural material is _____. <br> jade<br> quartz<br> diamond<br> talc
kakasveta [241]
Its diamond i am 100% right!

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To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you
Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log \frac{I}{I_o}

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

           β₂ = 10 log \frac{I_2}{I_o}

           

          β₂ -β₁ = 10 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

           \frac{I_2}{I_1} = 10³

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having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

           I = P / A

           P = I A

the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

          P =  I₁ A₁ =  I₂ A₂

          I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

         r₁² = 10³ r₂²

         

         r₂ = r₁ / √10³

         

we calculate

          r₂ = \frac{10.0}{\sqrt{10^3} }

          r₂ = 0.316 m

8 0
3 years ago
In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be
LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm  = m_2 * xcm

Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

                m_3 = 0.59 kg

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d1i1m1o1n [39]

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

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\omega = \frac{\theta}{t}

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now we have time period of the train given as

T = \frac{2\pi}{\omega}

T = \frac{2\pi}{\frac{\pi}{30}}

T = 60 s

5 0
4 years ago
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