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2 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]2 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]2 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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Rudik [331]

Answer:

I think is 2.

Explanation:

(The entire range of wavelengths or frequencies of electromagnetic radiation extending from gamma rays to the longest radio waves and including visible light)

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2 years ago

A batter hits a baseball with a bat. The bat exerts a force on the ball. Does the ball exert a force on the bat?

Mazyrski [523]

Yes, think about the difference of swinging a bat and not hitting a ball. It's fairly easy right? Now, when you hit a ball with the bat, you will feel the bat sting your hands. That's the force the ball is exerting on the bat!

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3 years ago

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A person walks 24 meters at 3 m/s. How long did it take them

asambeis [7]

Hi my dear friend,

2m*30 = 60m

~Thank you

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2 years ago

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Based on Archimedes' principle, the greatest buoyant force an object can experience in water is determined by which quantity?

kicyunya [14]

I think it the objects mass

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3 years ago

Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .