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Sergio [31]
3 years ago
7

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= f/m

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Hey guys can you help me solve this problem "how long will it take a car travelling 30m/s to come to stop ifs its acceleration i
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Answer:

10 seconds.

Explanation:

We can use a kinematic equation where we know the final velocity, initial velocity, acceleration, and need to determine the time <em>t: </em>

<em />\displaystyle v_f = v_i + at<em />

<em />

The initial velocit is 30 m/s, the final velocity is 0 m/s (as we stopped), and the acceleration is -3 m/s².

Substitute and solve for <em>t: </em>

<em />\displaystyle \begin{aligned} (0\text{ m/s}) & = (30 \text{ m/s}) + (-3 \text{ m/s$^2$}) t \\ \\  t & = \frac{-30\text{ m/s}}{-3 \text{ m/s$^2$}} \\ \\ & = 10 \text{ s} \end{aligned}<em />

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7 0
2 years ago
If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind.
AleksAgata [21]

Answer:

a) f_1=5.587Hz

b) f_{n+1}-f_n=5.587Hz

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The frequency of the n^{th} harmonic of a vibrating string of length <em>L, </em>linear density \mu under a tension <em>T</em> is given by the formula:

f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}

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f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz

b) The <em>frequency difference</em> between successive modes is the fundamental frequency, since:

f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz

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3 years ago
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displacement

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A. through a relatively short distance.

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