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EleoNora [17]
3 years ago
8

On a map of average annual temperature, why are the lower latitudes so much warmer than the higher latitudes?

Physics
2 answers:
Alenkasestr [34]3 years ago
3 0
Low altitudes are close to the equator. The temperatures are warmer as the latitude decreases, so they are very warm at low latitudes.

Readme [11.4K]3 years ago
3 0

Answer:

Explanation:

Weather is the combination of the parameters temperature, precipitation, wind, humidity, atmospheric pressure and cloudiness. These parameters are called weather elements. Climate factors are agents such as latitude, prevailing winds, sea currents, distance to the sea, altitude and relief, which modify, accentuate or limit the elements of the climate and give rise to different types of climates.

As mentioned, latitude is one of the factors that change the weather. Latitude is the angular distance from a point on the surface of the earth to the equator. It is measured in degrees, minutes and seconds.

The closer a place is to the equator, the solar energy will affect more vertically and with more intensity on it, so the warmer the temperatures will be. The further that point of the equator is found, the solar energy will reach it with a smaller angle. And if the point is near the poles, the sun's rays will arrive at a very small angle and only for a few months. This will make temperatures very low all year.

That is, in summary, the places that are at high altitudes (far from the equator) receive less sunlight than the places that are at low latitudes (near the equator) because the sun's rays have a greater or lesser obliqueness ( angle) over the earth's surface.

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34kurt
(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K)  I hoped that helped
4 0
3 years ago
A block weighs 15 n and is suspended from a spring that is attached to the ceiling. the spring stretches by 0.075 m from its uns
Illusion [34]

We can salve the problem by using the formula:

F=kx

where F is the force applied, k is the spring constant and x is the stretching of the spring.


From the first situation we can calculate the spring constant, which is given by the ratio between the force applied and the stretching of the spring:

k=\frac{F}{x}=\frac{15 N}{0.075 m}=200 N/m


By using the value of the spring constant we calculated in the first step, we can calculate the new stretching of the spring when a force of 33 N is applied:

x=\frac{F}{k}=\frac{33 N}{200 N/m}=0.165 m

4 0
4 years ago
1) Determine the magnitude of energy for each of the blanks on the diagram. Give the correct values for 1A, 1B, and 1C.
morpeh [17]

Answer:

Explanation: y’all taking the same test as me hahahahah I got the answers but I can’t attach the picture here so hit me up on snap daniela_0789

4 0
3 years ago
A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A
galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

8 0
3 years ago
A wind turbine is an example of what kind of device
Liula [17]

Answer:

it’s an example of a generator.

Explanation:

3 0
2 years ago
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