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Nikitich [7]
3 years ago
6

The intensity of electromagnetic waves from the sun is 1.4kW/m2 just above the earth's atmosphere. Eighty percent of this reache

s the surface at noon on a clear summer day. Suppose you think of your back as a 30.0 cm × 51.0 cm rectangle. How many joules of solar energy fall on your back as you work on your tan for 1.30 hr ?

Physics
2 answers:
zalisa [80]3 years ago
6 0

Explanation:

Below is an attachment containing the solution.

Arlecino [84]3 years ago
3 0

Answer:

E=8.02*10^{5}J

Explanation:

Given data

Electromagnetic waves from the sun is I=1.4kW/m² at 80%

Area a=(0.30×0.51)m²

Time t=1.30 hr

To find

Energy E

Solution

The energy received by your back is calculated as:

E=Pt=Iat\\E=(0.80)(1400W/m^{2} )(0.30*0.51m^{2} )(1.30*3600s)\\E=8.02*10^{5}J

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Vinvika [58]

Answer:

149,700,000 km   (= 1.50 x 10⁸ km)

Explanation:

Given,

Speed of light, c = 300,000 km/s

Time Taken = 8 min 19 s = 499 seconds

Recall, Distance = Speed x Time

= 300,000 km/s   x 499 s

= 149,700,000 km

= 1.50 x 10⁸ km

3 0
3 years ago
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The velocity of a particle traveling in a straight line is given by v = (6t - 3t²) m/s, where t is in seconds. If s = 0 when t =
qaws [65]

Answer

given,

v = (6 t - 3 t²) m/s

we know,

v = \dfrac{dx}{dt}

a = \dfrac{dv}{dt}

position of the particle

dx = v dt

integrating both side

\int dx = (6t - 3 t^2)\int dt

 x = 3 t² - t³

Position of the particle at t= 3 s

 x = 3  x 3² - 3³

 x = 0 m

now, particle’s deceleration

a = \dfrac{dv}{dt}

a = \dfrac{d}{dt}(6t - 3 t^2)

  a = 6 - 6 t

at t= 3 s

    a = 6 - 6 x 3

    a = -12 m/s²

distance traveled by the particle

  x = 3 t² - t³

at t = 0 x = 0

   t = 1 s   , x = 3 (1)² - 1³ = 2 m

   t = 2 s  ,  x = 3(2)² - 2³ = 4 m

   t = 3 s ,   x =  0 m

total distance traveled by the particle

D = distance in 0-1 s + distance in 1 -2 s + distance in 2 -3 s

D = 2 + 4 + 2 = 8 m

average speed of the particle

v_{avg} = \dfrac{distance}{time}

v_{avg} = \dfrac{8}{3}

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8 0
3 years ago
An object becomes positively charged when it
Montano1993 [528]

Answer:

D

Explanation:

8 0
4 years ago
A student removes a 10.5 kg stereo amplifier from a shelf that is 1.82 m high. The amplifier is lowered at a constant speed to a
Nonamiya [84]

Answer:

(a) the work done by the student is 110.1 J

(b) The gravitational force that acts on the amplifier is 102.9 N

Explanation:

Given;

mass of the amplifier, m = 10.5 kg

initial position of the amplifier, x₀ = 1.82 m

final position of the amplifier, x₁ =0.75 m

The dispalcement of the amplifier Δx = x₁ - x₀ = 1.82 m - 0.75 m = 1.07 m

(b) The gravitational force that acts on the amplifier;

F = mg

F = 10.5 x 9.8

F = 102.9 N

(a) the work done by the student is calculated as;

W = FΔx

W = 102.9 x 1.07

W = 110.1 J

7 0
3 years ago
Yassine on the football team tries to kick a football so that it stays in the air for a long "hang time". If the ball is kicked
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A is the answer sob I am sorry to bother
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