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natka813 [3]
3 years ago
13

Find the intensity of electromagnetic radiation at the surface of the sun (radius r=R=6.96×105kmr=R=6.96×105km). Ignore any scat

tering of the waves as they move radially outward from the center of the sun.
Physics
1 answer:
alisha [4.7K]3 years ago
7 0

Answer:

I = 4.46*10^{16}W/m^2.

Explanation:

Intensity I of the electromagnetic radiation is given by

I = \dfrac{P}{4\pi r^2},

where r is the distance from the EM source (the center of the sun, in our case), and P is the power output of the sun and it has the value

P = 3.9 *10^{26}W.

Since the radius of the sun in meters is r = 6.96*10^8km, the intensity I of the electromagnetic radiation at the surface of the sun is

I = \dfrac{3.9*10^{26}W}{4\pi (6.96*10^8m)^2}\\\\\boxed{ I = 4.46*10^{16}W/m^2}

The intensity of the electromagnetic radiation at the surface of the sun is I = 4.46*10^{16}W/m^2.

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A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark
Natali5045456 [20]

The total displacement of the toy car at the given positions is 0.

The given parameters;

  • <em>First displacement of the car, = 5 cm left</em>
  • <em>Second displacement of the car, = 8 cm right</em>
  • <em>Third displacement of the car, = 3 cm to the left</em>

The total displacement of the car is calculated as follows;

  • <em>Let the </em><em>left </em><em>direction be "</em><em>negative </em><em>direction"</em>
  • <em>Let the </em><em>right </em><em>direction be "</em><em>positive </em><em>direction"</em>

\Delta x = - \ 5\ cm  \ + \ 8 \ cm \ - \ 3 \ cm \\\\\Delta x = 0

Thus, the total displacement of the toy car at the given positions is 0.

Learn more about displacement here: brainly.com/question/18158577

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A 86g ball is dropped vertically to the floor from a height of 2.87m and bounces to a height of 1.28. What is the magnitude of t
irga5000 [103]

Answer:

The impulse received by the ball from the floor during the bounce is approximately 1.11329438 m·kg/s

Explanation:

The given mass of the ball, m = 86 g = 0.089 kg

The height from which the ball is dropped, H = 2.87 m

The height to which the ball bounces, h = 1.28 m

Mathematically, we have;

Δp = F·Δt

Where;

Δp = The change in momentum = m·Δv

F = The applied force

Δt = The time of contact with the force

The velocity of the ball just before it touches the ground, v₁ = -√(2·g·H)

The velocity with which the ball leaves, v₂ = √(2·g·h)

The change in momentum, Δp = m·(v₂ - v₁)

∴ Δp = m·(√(2·g·h) - (-√(2·g·H))) = m·(√(2·g·h) +√(2·g·H) )

The impulse, Δp, received by the ball from the floor during the bounce is given as follows;

Δp = 0.089 kg × (√(2 × 9.8 m/s² × 1.28 m) + √(2 × 9.8 m/s² × 2.87 m)) ≈ 1.11329438 m·kg/s

The impulse received by the ball from the floor during the bounce, Δp ≈ 1.11329438 m·kg/s

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