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Zolol [24]
4 years ago
6

We know that the earth's axis is tilted 23 ½ degrees. On or about June 21 or 22 each year, the summer solstice occurs for those

of us who live in the Northern Hemisphere. On this particular day, the sun's rays are striking earth directly, or at 90 degrees, at a latitude approximately 23 ½ degrees north of the equator. If you were at 23 1/2 degrees north latitude on that day, and you looked up at noon, the sun would be directly overhead.
Imagine you have taken a flight north to New York City. Describe the sun rays at exactly noon on the same day of the year.

A) The sun's rays would also be directly overhead. The sun's rays strike earth at 90 degrees throughout the hemisphere.

B) The sun's rays would also be directly overhead but only for one hour, twelve noon until one o'clock. After that the earth's movement distorts the angle of the sun's rays.

C) The sun's rays will be directly overhead in the Northern Hemisphere during the fall equinox. The movement of earth changes the angle of incidence of the sun's rays throughout the year.

D) The sun's rays will never be directly overhead. The latitude of 23 ½ degrees north is known as the Tropic of Cancer. Above this imaginary line the sun's rays hit earth with decreased angles.
Physics
2 answers:
Dima020 [189]4 years ago
8 0
<span>D) The sun's rays will never be directly overhead. The latitude of 23 ½ degrees north is known as the Tropic of Cancer. Above this imaginary line the sun's rays hit earth with decreased angles.</span>
arlik [135]4 years ago
8 0

Answer:

Option (D)

Explanation:

The path in which the sun moves across the sky usually varies from one season to another. This path is directly related to the position of sunrise and sunset, as the position of sunrise and sunset changes with the changing position of the sun. Even though the sun appears to be at the highest position during the winter solstice where the sun is at the maximum height and receives the highest amount of daylight, the sun is never directly overhead in the state of New York.

Thus, the rays of the sun never falls directly overhead in this region. At this condition, i.e above the tropic of cancer, the sun's rays hit the earth with gradually decreased angles.

Hence, the correct answer is option (D).

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A force of 200 N is applied at piston A, which has a surface area of 4
stira [4]

Answer:

60 cm²

Explanation:

Pressure = Force / Surface area

This is a 2 stage problem which requires working out pressure of piston A and then using that pressure from piston A to calculate the surface area of piston B

Firstly work out pressure of Piston A:

Pressure = 200 / 4

Pressure = 50 N/cm²

Then work out the Surface area of piston B :

Since the pressure is transferred the pressure stays the same so :

Surface area = Force / Pressure

Surface area = 3000 / 50

Surface area = 60 cm²

7 0
4 years ago
Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star.
Vsevolod [243]

Answer:

The angular speed of the neutron star is 3130.5 rad/s.

Explanation:

Given that,

Initial radius r_{1}=7\times10^{5}\ km

Final radius r_{2}=18 km

Density of a neutron \rho= 10^{14}

Equal masses of two stars m_{1}=m_{2}

Suppose, If the original star rotated once in 35 days, find the angular speed of the neutron star

Time period of original star T =  35 days = 3024000 s

We need to calculate the initial angular speed of original star

Using formula of angular star

\omega=\dfrac{2\pi}{T}

Put the value into the formula

\omega_{1}=\dfrac{2\pi}{3024000}

\omega_{1}=0.00207\times10^{-3}\ rad/s

Let the initial moment of inertia of the star is

I_{1}=m_{1}r_{1}^2

Final moment of inertia of the star is

I_{2}=m_{2}r_{2}^2

From the conservation of angular momentum

I_{1}\omega_{1}=I_{2}\omega_{2}

\omega_{2}=\dfrac{I_{1}\omega_{1}}{I_{2}}

\omega_{2}=\dfrac{m_{1}r_{1}^2\omega_{1}}{m_{2}r_{2}^2}

Put the value into the formula

\omega_{2}=\dfrac{(7.0\times10^{5})^2\times0.00207\times10^{-3}}{18^2}

\omega_{2}=3130.5\ rad/s

Hence, The angular speed of the neutron star is 3130.5 rad/s.

8 0
3 years ago
Can someone pls help with the last one
Leviafan [203]

Answer:

ummmmmmmmmmmm ask your teacher or perant for help

Explanation:

so umm ask an adult or teen figure and maybe you will get it write not sure do

6 0
3 years ago
A pole that is 2.9m tall casts a shadow that is 1.32m long. at the same time, a nearby tower casts a shadow that is 37.75m long.
Mrrafil [7]
\frac{2.9}{1.32}   = \frac{x}{37.75}

x = 82.93 m

Nothing to deal with physics.
Maths. Scale factor. Year 9.
5 0
3 years ago
A student walked 2km in .5 hours what is his average speed on the way to school
ddd [48]
Speed = Distance%Time

Speed is 0,4km/h


5 0
3 years ago
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