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3 years ago

How does the position of the Sun in the sky affect the intensity of sunlight striking Earth's surface?

Physics

2 answers:

Goshia [24]3 years ago

6 0

Answer:

First, the surface that the sun faces is almost always the same, but it changes the "position" during the lapse of a day.

Now, if we have a fixed position, as the day phases the effective area that the sun "sees" changes.

For example at the beginning of the day, the sun only sees an ellipse, when the sun is right above the area, the sun sees a surface like a circle, and as the sun starts to go away the surface turns again into an ellipse.

you can see this as holding a ball with a region painted on it and start to rotate the ball, you will see that in some positions you will not see the painted region, and as you rotate it you will start seeing a little part, then the whole part, then a little part again, and so on, where in this case your vision represents the sunlight, and the painted region is the fixed area of the planet.

evablogger [386]3 years ago

3 0

The Sun's position has a great impact on how much light reaches the surface. Such as at noon, the Sun has the least amount of gas light has to pass through. Meanwhile at dawn/dusk, it has a thicker layer of gas the light has to puncture.

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A 60-Hz single-phase transformer with capacity of 150 kVA has the following parameters: RP = 0.35 Ω RS = 0.002 Ω Rc = 5.2 kΩ XP

worty [1.4K]

Answer:

Explanation:

Total Resistance, Rt = (0.35 + 0.002 + 5200 + 0.5 + 0.008 + 1100) = 6300 ohms

a) Capacity of transformer, Pt = 150KVA = 150,000 W

Input Voltage, Vp = 2.8 KV = 2800 V

Current, Ip = 150000/2800 = 53.57 A

Input impedance, Zp = Vp/Ip = 2800/5357 = 52.27 ohms

b) i) Input current = 53.57 A

ii) Voltage, V = Ip * Rt = 53.57 X 6300 = 337.5 KV

iii) Power, P = I² * Rt = (53.57)² X 6300 = 18.08 MW

iv) Power factor = 0.83

8 0

2 years ago

An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid

Setler79 [48]

Answer:

$- 7.04\times10^{11} J$

$5.03\times10^{5} N$

Explanation:

$m$ = mass of the asteroid = 43000 kg

$v_{o}$ = initial speed of asteroid = 7600 m/s

$v$ = final speed of asteroid = 5000 m/s

$W$ = Work done by the force on asteroid

Using work-change in kinetic energy theorem

$W = (0.5) m (v^{2} - v_{o}^{2})\\W = (0.5) (43000) (5000^{2} - 7600^{2})\\W = - 7.04\times10^{11} J$

$F$ = magnitude of force on asteroid

$d$ = distance traveled by asteroid while it slows down = 1.4 x 10⁶ m

Work done by the force on the asteroid to slow it down is given as

$W = - F d\\- 7.04\times10^{11} = - F (1.4\times10^{6})\\F = 5.03\times10^{5} N$

4 0

2 years ago

An eagle is flying horizontally at a speed of 3.10 m/s when the fish in her talons wiggles loose and falls into the lake 6.10 m

Alinara [238K]

Answer:

10.93m/s with the assumption that the water in the lake is still (the water has a speed of zero)

Explanation:

The velocity of the fish relative to the water when it hits the water surface is equal to the resultant velocity between the fish and the water when it hits it.

The fish drops on the water surface vertically with a vertical velocity v. Nothing was said about the velocity of the water, hence we can safely assume that the velocity if the water in the lake is zero, meaning that it is still. Therefore the relative velocity becomes equal to the velocity v with which the fish strikes the water surface.

We use the first equation of motion for a free-falling body to obtain v as follows;

v = u + gt....................(1)

where g is acceleration due to gravity taken as 9.8m/s/s

It should also be noted that the horizontal and vertical components of the motion are independent of each other, hence we take u = 0 as the fish falls vertically.

To obtain t, we use the second equation of motion as stated;

$h=ut+gt^2/2.................(2)$