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Varvara68 [4.7K]
3 years ago
6

Only about 50% of the solar energy directed toward Earth penetrates directly to the surface. Explain what happens to the rest of

the radiation.
Physics
2 answers:
Hoochie [10]3 years ago
6 0

Answer:

A. It is absorbed or reflected by the atmosphere.

Explanation:

I did the test

Tresset [83]3 years ago
4 0

Answer:

it is absorbed or reflected by the atmosphere

Explanation:

In the case when approx 50% only of the solar energy would be directed towards earth and it would be penetrates directly to the surface so the rest or remaining of the radiation would be either absorbed or refected by the atmosphere

So as per the given situation the above represent the answer

hence, the same is to be considered and relevant

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An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below
grin007 [14]
<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let P be the pressure at a point.

Let p be the density fluid at a point.

Let v be the velocity of fluid at a point.

Bernoulli's equation states that P+\frac{1}{2}pv^{2}+pgh=constant for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let p_{up} be the pressure of a point just above the wing.

Let p_{do} be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}

So,p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa

Force is given by the product of pressure difference and area.

Given that area is 27ms^{2}.

So,lifting force is 16398.48\times 27=442758.96N

6 0
3 years ago
A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of
Vladimir79 [104]

Answer:

m = 2.23 \times 10^{-32} kg

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

orbital speeds = 220 km/s

we know that

acceleration due to centripetal force isa =   \frac{F}{m} = \frac{V^2}{r}

Gravitational forceF= \frac{Gm m}{d^2}

we know that

v = \frac{2\pi R}{T}

solving for

R = \frac{vT}{2\pi}

F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}

F = G\times \frac{\pi m}{(vT)^2}

a = \frac{v^2}{\frac{vT}{2\pi}}

a = \frac{2\pi v}{T}

we know that

f =ma

G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}

solving for m

m = \frac{2Tv^3}{\pi G}

m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}

m = 2.23 \times 10^{-32} kg

5 0
3 years ago
A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re
slamgirl [31]

Answer:

  • tension: 19.3 N
  • acceleration: 3.36 m/s^2

Explanation:

<u>Given</u>

  mass A = 2.0 kg

  mass B = 3.0 kg

  θ = 40°

<u>Find</u>

  The tension in the string

  The acceleration of the masses

<u>Solution</u>

Mass A is being pulled down the inclined plane by a force due to gravity of ...

  F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

  F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

  F/m = a = F/m

  (T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

  T = (2(29.4) +3(12.5986))/5 = 19.3192 N

__

Then the acceleration of B is ...

  a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0
3 years ago
The magnetic field 0.0 2M from a wire is 0.1 T. What is the magnitude of the magnetic field of 0.0 1M from the same wire?
Anika [276]

Answer:

Explanation:

Question

5 0
2 years ago
A 65kg person throws a 0.045kg snowball forward with a ground speed of 30m/s. A second person, with a mass of 60kg, catches the
asambeis [7]

Answer:

v₁ = 2.48m/s, v₂ = 0.02m/s

Explanation:

Momentum p must be conserved. p = mv

1) First person throwing the snow ball. The momentum before the throw:

p = (65kg + 0.045kg) * 2.5 m/s

The momentum after the throw:

p = 65kg * v₁ + 0.045kg * 30m/s

Solving for the velocity v₁ of person 1:

v₁ = ((65kg + 0.045kg) * 2.5 m/s - 0.045kg * 30m/s) / 65kg = 2.48m/s

2) Second person catching the ball. The momentum before the catch:

p = 0.045kg * 30m/s + 60kg * 0m/s

The momentum after the catch:

p = (60kg + 0.045kg) * v₂

Solving for velocity v₂ of person 2:

v₂ = 0.045kg * 30m/s / (60kg + 0.045kg) = 0.02 m/s

5 0
3 years ago
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