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zalisa [80]
3 years ago
12

On the average, about what percentage of the solar energy that strikes the outer atmosphere eventually reaches the earth's surfa

ce
Physics
1 answer:
almond37 [142]3 years ago
5 0
I think the answer to your question is twenty-four percent.
You might be interested in
Bicycle A with mass 40 Kg is traveling with a velocity of 4 m/s and Bicycle B with mass 20 Kg is traveling with a velocity of 2m
andrew-mc [135]

Answer:

bicycle A has a greater K.E.

Explanation:

K.E = 1/2mv²

bicycle A = 1/2 × 40 × 4² = 320J

bicycle B = 1/2 × 20 × 2² = 100J

bicycle A has a greater K.E. because it has bigger mass and moves with faster velocity

7 0
3 years ago
Students are going to conduct an experiment to study the effect of a net force applied to an object on the object’s motion. In e
Ronch [10]

Answer:

A. velocity and time

Explanation:

A force can be define as an agent which has the capacity to change the state of an object. It can either increase the velocity of a body, change its direction of motion or cause a moving object to come to rest.

From Newton's second law of motion;

F = ma

where F is the force on the object, m is the mass of the object and a is the acceleration of the object. The unit of force is kgm/s^{2} or Newtons.

a = \frac{change in velocity}{change in time}

In the given question, apart from the mass of the object which is constant, the students should take the measurements of the velocity and time in each trial so as to calculate the required acceleration.

7 0
3 years ago
Read 2 more answers
Not only did Skid get shot out of a cannon when he was a clown in the circus, but he was also launched into the air by a vertica
tiny-mole [99]

Answer:

v = 16.11 m / s

Explanation:

For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation

starting point. When the spring is compressed

        Em₀ = K_e + U = ½ k x² + m g x ’

final point. The point where it leaves the platform

        Em_f = K = ½ m v²

energy is conserved

         Em₀ = Em_f

         ½ k x² + m g x ’= ½ m v²

         v² = \frac{k}{m}  x² + g x

let's calculate

         v² = \frac{8700}{55}  1.25² + 9.8 1.25

         v² = 247.159 + 12.25 = 259.409

         v = 16.11 m / s

8 0
3 years ago
Pluto is now classified as a ________ in our solar system.
Sonbull [250]
Pluto is classified as a dwarf planet!
5 0
3 years ago
Read 2 more answers
Suppose that 4.4 moles of a monatomic ideal gas (atomic mass = 7.9 × 10-27 kg) are heated from 300 K to 500 K at a constant volu
timurjin [86]

Answer:

1) ΔQ₁ = 10.97 x 10³ J = 10.97 KJ

2) W₁ = 0 J

3) P = 41.66 x 10³ Pa = 41.66 KPa

4) v = 1618.72 m/s

5) ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ

6) W₂ = - 7.33 KJ

Explanation:

1)

The heat transfer for a constant volume process is given by the formula:

ΔQ₁ = ΔU = n Cv ΔT

where,

ΔQ₁ = Heat transfer during constant volume process

ΔU = Change in internal energy of gas

n = No. of moles = 4.4 mol

Cv = Molar Specific Heat at Constant Volume = 12.47 J/mol.k

ΔT = Change in Temperature = T₂ - T₁ = 500 k - 300 k = 200 k

Therefore,

ΔQ₁ = (4.4 mol)(12.47 J/mol.k)(200 k)

<u>ΔQ₁ = 10.97 x 10³ J = 10.97 KJ</u>

<u></u>

2)

Since, work done by gas is given as:

W₁ = PΔV

where,

ΔV = 0, due to constant volume

Therefore,

<u>W₁ = 0 J</u>

<u></u>

4)

The average kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

but, K.E is also given by:

K.E = (1/2)mv²

Comparing both equations:

(1/2)mv² = (3/2)KT

mv² = 3KT

v = √(3KT/m)

where,

v = average speed of gas molecue = ?

K = Boltzman Constant = 1.38 x 10⁻²³ J/k

T = Absolute Temperature = 500 K

m = mass of a molecule = 7.9 x 10⁻²⁷ kg

Therefore,

v = √[(3)(1.38 x 10⁻²³ J/k)(500 k)/(7.9 x 10⁻²⁷ kg)]

<u>v = 1618.72 m/s</u>

<u></u>

3)

From kinetic molecular theory, we know that or an ideal gas:

P = (1/3)ρv²

where,

P = pressure of gas = ?

m = Mass of Gas = (Atomic Mass)(No. of Atoms)

m = (Atomic Mass)(Avogadro's Number)(No. of Moles)

m = (7.9 x 10⁻²⁷ kg/atom)(6.022 x 10²³ atoms/mol)(4.4 mol)

m = 0.021 kg

ρ = density = mass/volume = 0.021 kg/0.44 m³ = 0.0477 kg/m³

Therefore,

P = (1/3)(0.0477 kg/m³)(1618.72 m/s)²

<u>P = 41.66 x 10³ Pa = 41.66 KPa</u>

<u></u>

5)

The heat transfer for a constant pressure process is given by the formula:

ΔQ₂ =  n Cp ΔT

where,

ΔQ₂ = Heat transfer during constant pressure process

n = No. of moles = 4.4 mol

Cp = Molar Specific Heat at Constant Pressure = 20.79 J/mol.k

ΔT = Change in Temperature = T₂ - T₁ = 300 k - 500 k = -200 k

Therefore,

ΔQ₂ = (4.4 mol)(20.79 J/mol.k)(-200 k)

<u>ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ</u>

<u>Negative sign shows heat flows from system to surrounding.</u>

<u></u>

6)

From Charles' Law, we know that:

V₁/T₁ = V₂/T₂

V₂ = (V₁)(T₂)/(T₁)

where,

V₁ = 0.44 m³

V₂ = ?

T₁ = 500 K

T₂ = 300 k

Therefore,

V₂ = (0.44 m³)(300 k)/(500 k)

V₂ = 0.264 m³

Therefore,

ΔV = V₂ - V₁ = 0.264 m³ - 0.44 m³ = - 0.176 m³

Hence, the work done , will be:

W₂ = PΔV = (41.66 KPa)(- 0.176 m³)

<u>W₂ = - 7.33 KJ</u>

<u>Negative sign shows that the work is done by the gas</u>

4 0
3 years ago
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