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Sindrei [870]
3 years ago
8

How do lines of latitude affect how direct or indirect the Sun’s rays are on the Earth?

Physics
1 answer:
IgorLugansk [536]3 years ago
8 0
The technical definition of latitude is the angular distance north or south from the earth's equator measured through 90 degrees. ... Locations at lower latitudes receive stronger and more direct sunlight than locations near the poles. Energy input from the sun is the main driving force in the atmosphere.



The Seasons at Different Latitudes
The seasonal effects are different at different latitudes on Earth. Near the equator, for instance, all seasons are much the same. Every day of the year, the Sun is up half the time, so there are approximately 12 hours of sunshine and 12 hours of night.



When we consider Latitude alone as a control, we know that the low latitudes (say from the Equator to approximately 30 degrees N/S) are the warmest across the year (on an annual basis).
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3 years ago
Find the velocity of a water wave in meters per second with frequency
Yuri [45]

Answer:

Velocity, v = 0.164 m/s

Explanation:

We have,

Frequency of wave, f = 0.04 Hz

Wavelength, \lambda=4.1\ m

It is required to find the velocity of a water wave. The speed of any wave is given in terms of wavelength and frequency. Its formula is :

v=f\lambda\\\\v=0.04\times 4.1\\\\v=0.164\ m/s

So, the velocity of a water wave is 0.164 m/s.

7 0
3 years ago
Determine the gravitational potential energy, in kJ, of 3 m3 of liquid water at an elevation of 40 m above the surface of Earth.
melomori [17]

Explanation:

We will calculate the gravitational potential energy as follows.

                 P.E_{1} = mgz_{1}

       P.E_{1} = (\rho V)gz_{1}    

                    = 1000 kg/m^{3} \times 3 m^{3} \times 9.7 \times 40 m

                    = 1164000 J

or,                = 1164 kJ         (as 1 kJ = 1000 J)

Now, we will calculate the change in potential energy as follows.

             \Delta P.E = mg(z_{2} - z_{1})

                         = \rho \times V \times g (z_{2} - z_{1})

                         = 1000 \times 3 \times 9.7 (10 - 40)m

                         = -873000 J

or,                      = -873 kJ

Thus, we can conclude that change in  gravitational potential energy is -873 kJ.

4 0
2 years ago
What is the 5 sources of chemical weathering
irakobra [83]
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Hope this helps.
6 0
3 years ago
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Alexeev081 [22]

Now let’s say you’re on the Moon. If you were to drop a hammer and a feather from the same height, which would hit the ground first?

Trick Question! On the moon both objects would hit the ground at the same time. On Earth, the hammer lands first.

So yeah, the student is right. Galileo gave us this theory long ago.

5 0
2 years ago
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