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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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Sunny_sXe [5.5K]

Answer:

0.5 , 54.5

Explanation:

for acceleration we should derivate the equation 2 times

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To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(
NeX [460]

Answer:

The transverse displacement is   y(1.51 , 0.150) = 0.055 m    

Explanation:

 From the question we are told that

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     The speed of the transverse wave is v = 8.25 \ m/s

     The amplitude of the transverse wave is A = 5.50 *10^{-2} m

     The wavelength of the transverse wave is \lambda = 0540 m

      At t= 0.150s , x = 1.51 m

 The angular frequency of the wave is mathematically represented as

          w = vk

Substituting values  

         w = 8.25 * 11.64

        w = 96.03 \ rad/s

The propagation constant k is mathematically represented as

                  k = \frac{2 \pi}{\lambda}

Substituting values

                  k = \frac{2 * 3.142}{0. 540}

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Substituting values into the equation for mechanical waves

      y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) - (96.03  * 0.150))

       y(1.51 , 0.150) = 0.055 m    

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Answer:

Explanation:

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natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

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and

\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

e = \sqrt{\frac{h_1}{6.1} }

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- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

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3 years ago
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