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3 years ago

Describe the cyclic patterns of lunar phases and eclipses of the Sun and moon.

1 answer:

4 0

Cyclic Patterns of Lunar Phases, Eclipses of the Sun and Moon, and Seasons. Waning Gibbous is the size between a Full Moon and a Third Quarter, and means three fourths of a full moon, but getting smaller.
Our Moon's shape doesn't really change — it only appears that way! The “amount” of Moon that we see as we look from Earth changes in a cycle that repeats about once a month (29.5 days). The relative positions of our Sun, Earth, and Moon, cause these changes.

As the Sun sets, the Moon rises with the side that faces Earth fully exposed to sunlight (5).The Moon has phases because it orbits Earth, which causes the portion we see illuminated to change. The Moon takes 27.3 days to orbit Earth, but the lunar phase cycle (from new Moon to new Moon)

Hope
this helps
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You might be interested in

A material's ability to allow heat or electricity to flow through it...

NikAS [45]

Answer:

Explanation:

how well a material shines or reflects light

8 0

3 years ago

A tourist averaged 82km/hr for a 6.5h trip in her volkswagen. how far did she go?

Whitepunk [10]

82 ÷ 6.5 = 12.615384615384... repeating
round = 12.6
12.6 · 6.5 = 81.9
round = 82
She went an average of 12.6 km an hour
Hope this helps! ;)

4 0

3 years ago

Determine the values of m and n when the following average distance from the Sun to the Earth is written in scientific notation:

Leviafan [203]

Answer:

$150000000000\ m=1.5\times 10^{11}\ m$

Explanation:

A number can be written in the form of :

$N=m\times 10^n$

Where

m is the real number

n is any integer

In this case, the average distance from the Sun to the Earth is given,

d = 150000000000 meters

There are 10 zeroes in this number. We need to write this number in scientific notation. It is given by :

$d=1.5\times 10^{11}\ m$

So, the average distance from the Sun to the Earth is $d=1.5\times 10^{11}\ m$ . Hence, this is the required solution.

3 0

3 years ago

a roller coaster is at the top of a hill and rolls to the top of a lower hill.If mechanical energy is conserved,on the top of wh

pychu [463]

Hello!!

Here we have a simple matter of conservation of energy. ME=PE+KE.

At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.

Hope this helps you understand the concept!! Any questions please just ask!! Thank you so much!!

7 0

4 years ago

Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia

Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km west

velocity V1 = 8.9 km/h

distance B = 4.5 km east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7 ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x ) / 8.9 .............2

now equating equation 1 and 2

time A = time B

x / 7 = ( 10.2 - x ) / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0

3 years ago