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r-ruslan [8.4K]
2 years ago
8

Why is the Earth generally warmer at the equator than at the poles?

Physics
2 answers:
Alchen [17]2 years ago
7 0

Answer: The Sun's energy is more concentrated on the equator than the poles. This is because the Sun's energy (in the form of warmth) is more direct at the equator, and less direct at the poles.

Explanation: You can see this with a flashlight. When you point the flashlight directly toward a tale, the light is very concentrated and right. However, when you point the flashlight toward the table at an angle, the light becomes more spread out and less concentrated.  This is why even though sometimes the North Poles and South Poles have 24 hours of sunlight, it never actually reaches a high temperature and instead, keeps it's cold temperature.

Elenna [48]2 years ago
5 0
Because of the earths curve and temperature. The temperature in an area depends on the amount of the Sun's energy reaching the surface in that area. The equator tilts closer to the sun than the poles do. I hope this helps!
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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude
Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

\omega_f = \omega_i + \alpha t

where

\omega_i = 0.300 rev/s is the initial angular velocity

\alpha = 0.895 rev/s^2 is the angular acceleration

Substituting t = 0.200 s, we find

\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s

Let's now convert it into rad/s:

\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

r=\frac{0.800}{2}=0.400 m

And so now we can find the tangential speed at t = 0.200 s:

v=\omega_f r =(3.01)(0.400)=1.2 m/s

2) 2.25 m/s^2

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

a_t = \alpha r

where \alpha is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

a_t = (5.62)(0.400)=2.25 m/s^2

3) 3.6 m/s^2

The centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

a=\frac{1.2^2}{0.400}=3.6 m/s^2

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3 years ago
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