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2 years ago

11

Monochromatic light is incident on a metal surface, and electrons are ejected. If the intensity of the light increases, what wil

l happen to the ejection rate of the electrons?

1 answer:

drek231 [11]2 years ago

4 0

Answer:The rate of ejection of photoelectrons will increase

Explanation:

If the frequency of incident monochromatic light is held constant and its intensity is increased, the rate of ejection of photoelectrons from the metal surface increases with increase in intensity of the monochromatic light. More current flows due to more ejection of photoelectrons.

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The odometer of a car changes from 1048 km to 1096 km in 40

Makovka662 [10]

Answer:

20m/s

Explanation:

it covers 20 metres in a second

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1 year ago

One cereal bar has a mass of 37 g. What is the mass of 6 cereal bars? Is that more than or less than 1 kg? Explain your answer.

GuDViN [60]

The mass is 222g. No, it is less than 1kg. There are 1000 grams in a kilogram so it would be 0.222kg. Hope this Helps :D

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3 years ago

A theory is a hyothesis that has been varified by multiple investigations. <br> True or false

irakobra [83]

<span>A theory is a hyothesis that has been varified by multiple investigations.

true</span>

7 0

2 years ago

A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when

Marianna [84]

Answer:

A) $P=13.92\ J.s^{-1}$

B) $v=3730.9912\ m.s^{-1}$

C) $v=74.44\ mm.s^{-1}$

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

Distance form the sound source, $s=50\ m$

sound intensity level at the given location, $\beta=114\ dB$

diameter of the eardrum membrane in humans, $d=8.4 \times 10^{-3}\ m$

We have the minimum detectable intensity to the human ears, $I_0=10^{-12}\ W.m^{-2}$

(A)

<u>Now the intensity of the sound at the given location is related mathematically as:</u>

$\beta=10\ log(\frac{I}{I_0} )$ ..........................................(1)

$114=10\ log\ (\frac{I}{10^{-12}} )$

$11.4=log\ I+12\ log\ 10$

$I=0.2512\ W.m^{-2}$

<em>As we know :</em>

$I=\frac{P}{A}$

$0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }$

$P=13.92\ J.s^{-1}$ is the energy transferred to the eardrums per second.

(B)

mass of mosquito, $m=2\times 10^{-6}\ kg$

<u>Now the velocity of mosquito for the same kinetic energy:</u>

$KE=\frac{1}{2} m.v^2$

$13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2$

$v=3730.9912\ m.s^{-1}$

(C)

Given:

Sound intensity, $\beta = 20\ dB$

<u>Using eq. (1)</u>

$20=10\ log\ (\frac{I}{10^{-12}} )$

$2=log\ I+12\ log\ 10$

$I=10^{-10}\ W.m^{-2}$

Now, power:

$P=I.A$

$P=10^{-10}\times \pi\times \frac{8.4^2}{4}$

$P=5.54\times 10^{-9}\ J.s^{-1}$

Hence:

$KE=\frac{1}{2} m.v^2$

$5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2$

$v=0.07444\ m.s^{-1}$

$v=74.44\ mm.s^{-1}$

(D)

mosquitoes speed in part B is very much larger than that of part C.

7 0

2 years ago

If you are driving 90 km/hkm/h along a straight road and you look to the side for 2.8 ss , how far do you travel during this ina

pashok25 [27]

Answer:

Distance = 70 meters

Explanation:

<u>Given the following data;</u>

Speed = 90 km/h

Time = 2.8 seconds

<u>Conversion:</u>

90 km/h to meters per seconds = 90 * 1000/3600 = 90000/3600 = 25 m/s

To find the distance covered during this inattentive period;

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the formula;

$Speed = \frac{distance}{time}$

Making distance the subject of formula, we have;

$Distance = speed * time$

Substituting into the above formula, we have;

$Distance = 25 * 2.8$

<em>Distance = 70 meters</em>

5 0

2 years ago