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amm1812
3 years ago
8

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

Physics
1 answer:
svlad2 [7]3 years ago
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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Two identical charged pith balls are brought together to touch each other. They are then
sergejj [24]

Answer:

-17.5 nC

Explanation:

charge A = -30 nC

charge B = -5 nC

After adding them it would be the average of the two charges because of the getting same voltage difference. so

c = (-30+(-5)) / 2 nC

c= -17.5 nC

answer is -17.5 nC

4 0
3 years ago
Can you die if you bust a hemorrhoid?
SVEN [57.7K]
I think that you can die
8 0
3 years ago
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A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find th
Rashid [163]

Answer:

The velocity of the particle = -1.92 m/s

The speed of the particle = 5.72 m/s

Explanation:

Given equation of motion;

f(t) = 18 \ + \ \frac{48}{t} \ + \ 1

Velocity is defined as the change in displacement with time.

V = \frac{df(t)}{dt} = -\frac{48}{t^2} \\\\at \ t = 5 \ s\\\\V = -\frac{48}{5^2} = \frac{-48}{25} = - 1.92 \ m/s

The distance traveled by the particle in 5 s:

s = f(5) = 18 + \frac{48}{5} + 1\\\\s= 28.6 \ m

The speed of the particle when t = 5s

Speed = \frac{28.6}{5} = 5.72 \ m/s

6 0
3 years ago
A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?
Olenka [21]
Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m

Formula:
F = k*x

Solving: 
F = k*x
F = 50*0.15
\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark

Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
E = \frac{k*x^2}{2}

Solving:(Energy associated with this stretching)
E = \frac{k*x^2}{2}
E =  \frac{50*0.15^2}{2}
E =  \frac{50*0.0225}{2}
E =  \frac{1.125}{2}
\boxed{\boxed{E = 0.5625\:J}}\end{array}}\qquad\quad\checkmark

7 0
3 years ago
Please help with this, oh and bungee gum contains both the properties of rubber AND gum you see?
Sladkaya [172]
Can I still get 5 points bc u already figured it out
8 0
3 years ago
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