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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. a gradual approximation to the final solution

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The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m
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A) The resultant force is 30.4 N at 25.3^{\circ}

B) The resultant force is 18.7 N at 43.9^{\circ}

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

F_1 = 20 N at 0^{\circ} above x-axis

F_2 = 15 N at 60^{\circ} above y-axis

Resolving each force:

F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N

F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N

So, the components of the resultant are:

F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N

And the magnitude of the resultant is:

F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N

And the direction is:

\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

\theta=180^{\circ}-60^{\circ}=120^{\circ}

So we have:

F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N

So, the components of the resultant this time are:

F_x = F_{1x}+F_{2x}=20-7.5 = 12.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N

And the magnitude is:

F=\sqrt{F_x^2+F_y^2}=\sqrt{13.5^2+13.0^2}=18.7 N

And the direction is:

\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}

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7 0
3 years ago
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