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kirza4 [7]
3 years ago
8

When a beam of light is incident on a surface, it delivers energy to the surface. The intensity of the beam is defined as the en

ergy delivered per unit area per unit time.
What is the unit of intensity, expressed in SI base units?​
Physics
1 answer:
scoundrel [369]3 years ago
6 0

Answer:

Kg

Explanation:

Given that the definition of intensity is:

The intensity of the beam is defined as the energy delivered per unit area per unit time. That is,

I =( E × t)/A

Which can expressed as

I = J/m^2/s

The S.I Units of

Energy is Joule = kgm^2s^-1

Area A = m^2

Time t = s

Intensity I = kgm^2s^-1 × m^-2 × s

Intensity I = kg

The cubic metres and seconds are cancelled out

Therefore, the S.I units of intensity is kg

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A diver is on a board 1.80 m above
Virty [35]

Answer:

v = 6.95 m/s

Explanation:

Given that,

A diver is on a board 1.80 m above  the water, s = 1.8 m

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Let v is the speed with which she hit the water. It will move under the action of gravity. Using the equation of motion as follows :

v^2-u^2=2gs\\\\v=\sqrt{u^2+2gs} \\\\v=\sqrt{(3.62)^2+2(9.8)(1.8)} \\\\v=6.95\ m/s

So, she will hit the water with a speed of 6.95 m/s.

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Which data set has the largest range?
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A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant- frequency tone whe
joja [24]

Answer:

The Doppler Effect is given by the following relation;

f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

v_o = The velocity of the observer

v_s = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(ii) When the source is receding from the observer, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

(1) Given that (v + v_s) > v, therefore, v < (v + v_s), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(1) Given that (v - v_s) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively greater (higher)

Explanation:

7 0
3 years ago
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