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2 years ago

How changes in solar position influence the intensity of radiation on a horizontal surface?

1 answer:

8 0

Changes in solar position influence the intensity of radiation on a horizontal surface because Energy concentration is greatly affected by changes in solar position. Energy intensity is greater when solar position is high. This concentrates more energy per unit area.

What determines the intensity of solar radiation?

The angle of the sun determines solar irradiance. The greater the angle, the lower the solar intensity. The lower the angle of the sun, the larger amount of ozone the light has to pass through.

What affects the intensity of solar energy?

The main parameters affecting the intensity of solar radiation are the zenith angle zeta of the sun, the water vapor content w of the atmosphere, and Schupp's turbidity coefficient B.

To learn more about solar radiation here

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Two vectors of magnitudes 30 units and 70 units are added to each other. What are possible results of this addition? (section 3.

Yanka [14]

Answer:

the correct answer is option C which is 50 units.

Explanation:

given,

two vector of magnitude = 30 units and of 70 units

to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}

cos θ value varies from -1 to 1

so, resultant vector

= $\sqrt{a^2+b^2-2 a b cos\theta}\ to\ \sqrt{a^2+b^2+ 2 a b cos\theta}$

a = 30 units and b = 70 units

= $\sqrt{30^2+70^2-2\times 30\times 70}\ to\ \sqrt{30^2+70^2+2\times 30\times 70}$

= 40 units to 100 units

hence, the correct answer is option C which is 50 units.

4 0

3 years ago

Around which months does el nino end and la nina start?

ad-work [718]

Answer:

They both tend to develop during the spring (March-June), reach peak intensity during the late autumn or winter (November-February), and then weaken during the spring or early summer (March-June)

3 0

3 years ago

Read 2 more answers

A loading car is at rest on a track forming an angle of 25° with the vertical. The gross weight of the car and its load is 5500

koban [17]

Idk I just need other answers

4 0

3 years ago

Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed,

KATRIN_1 [288]

Answer:

a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

V = i (R₁ + R₂)

with R₁ connected

V = (i + 0.5) R₁

with R₂ connected

V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

V = i ( $\frac{V}{ i+0.5} + \frac{V}{i+0.25}$ )

1 = i ( $\frac{1}{i+0.5} + \frac{1}{i+0.25}$ )

1 = i ( $\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }$ ) = $\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}$

i² + 0.75 i + 0.125 = 2i² + 0.75 i

i² - 0.125 = 0

i = √0.125

i = 0.35355 A

with the second equation we look for R1

R₁ = $\frac{V}{i+0.5}$

R₁ = 12 /( 0.35355 +0.5)

R₁ = 14.1 Ω

with the third equation we look for R2

R₂ = $\frac{V}{i+0.25}$

R₂ = $\frac{12}{0.35355+0.25}$

R₂ = 19.9 Ω

3 0

2 years ago

Which of the following correctly compares the uses of bar graphs versus pie charts?

slavikrds [6]

Bar graphs show data involved in distinct categories that do not overlap; whereas, pie charts show data as parts out of a whole (such as out of 100%).

6 0

3 years ago