Because the Earth's axis is not "straight up and down" as we move
around the sun.
So when we're on one side of the sun, the top pole leans slightly toward
the sun. During that time the sun shines more directly on the top half
of the Earth, and less directly on the bottom half. The people on the
top half see the sun higher in the sky, and their weather is warmer,
while the people on the bottom half see the sun lower in the sky, and
their weather is cooler.
Then, when we're on the other side of the sun, the top pole leans slightly
away from the sun. During that time the sun shines more directly on the
bottom half
of the Earth, and less directly on the top half. The people on
the bottom half see the sun higher in the sky, and their weather is warmer,
while the people on the top half see the sun lower in the sky, and their
weather is cooler.
The Earth makes the complete trip around the sun in one year, so the
people on the Earth go through this cycle of higher/lower sun and
warmer/cooler weather every year.
Answer:
Hello your question is poorly written below is the complete question
Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?
answer :
a) 231.48 days
b) n = 3.125 * 10^15
Explanation:
Battery moved 10,000 coulombs
current rate = 0.5 mA
<u>A) Determine how long the clock run on the battery. use the relation below</u>
q = i * t ----- ( 1 )
q = charge , i = current , t = time
10000 = 0.5 * 10^-3 * t
hence t = 2 * 10^7 secs
hence the time = 231.48 days
<u>B) Determine how many electrons per second flowed </u>
q = n*e ------ ( 2 )
n = number of electrons
e = 1.6 * 10^-19
q = 0.5 * 10^-3 coulomb ( charge flowing per electron )
back to equation 2
n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )
hence : n = 3.125 * 10^15
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The potential energy of the object depends on
- the height of the object with respect to some reference points,
- the mass of the object,
- the gravitational field the object is in.
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Hope it helps ~
The centripetal acceleration a is 4.32 
10^-4 m/s^2.
<u>Explanation:</u>
The speed is constant and computing the speed from the distance and time for one full lap.
Given, distance = 400 mm = 0.4 m, Time = 100 s.
Computing the v = 0.4 m / 100 s
v = 4 10^-3 m/s.
radius of the circular end r = 37 mm = 0.037 m.
centripetal acceleration a = v^2 / r
= (4 10^-3)^2 / 0.037
a = 4.32 10^-4 m/s^2.
