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kipiarov [429]
3 years ago
11

What are the orders of the lunar cycle?

Physics
1 answer:
skad [1K]3 years ago
8 0
The 8 moon phases in order are New moon, Waxing Crescent, First Quarter, Waxing Gibbous, Full moon, Waning Gibbous, Third Quarter, and finally Waxing Crescent.
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A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force \vec{F} ​F ​⃗ ​​ so that the swing ropes
Virty [35]

Answer:

169.74 N

Explanation:

Given,

Mass of the girl = 30 Kg

angle of the rope with vertical, θ = 30°

equating the vertical component of the tension

vertical component of the tension is equal to the weight of the girl.

 T cos θ = m g

 T cos 30° = 30 x 9.8

 T = 339.48 N

Tension on the two ropes is equal to 339.48 N

Tension in each of the rope = T/2

                                           = 339.48/2 = 169.74 N

Hence, the tension in each of the rope is equal to 169.74 N

7 0
3 years ago
A chemist needs to order an element that will not react with any other element. Which element should he order
Gwar [14]

Answer:

Helium

Explanation:

Helium is the least reactive element, since it is a noble gas with the smallest amount of valence rings.

4 0
3 years ago
List Five examples from daily life in which you see periodic motion caused by a pendulum
Basile [38]

Answer:

by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave.

Explanation:

5 0
3 years ago
A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit
Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=8.7\cdot 10^{13}kg is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

m\frac{v^2}{r}=G\frac{Mm}{r^2}

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s

3 0
3 years ago
A boy got out of a boat and as he
emmainna [20.7K]

Answer:

third law of motion

Explanation:

6 0
3 years ago
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