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poizon [28]
3 years ago
14

Which option BEST describes the position of the Sun, the Moon, and Earth during the new moon phase and the MOST LIKELY effect of

the phase on the petrel?
A.

The Sun, the Moon, and Earth are in a straight line, with Earth between the Sun and the Moon. The petrel will be very active because there will be a lot of light reflected by the Moon.

B.

The Sun, the Moon, and Earth are in a straight line, with the Moon between the Sun and Earth. The petrel will be less active because there will not be any light reflected by the Moon.

C.

The Sun, the Moon, and Earth are in a straight line, with the Moon between the Sun and Earth. The petrel will be very active because there will be a lot of light reflected by the Moon.

D.

The Sun, the Moon, and Earth are in a straight line, with Earth between the Sun and the Moon. The petrel will not be very active because there will not be any light reflected by the Moon..
Physics
1 answer:
artcher [175]3 years ago
3 0
A!!!!!!!!!!!!!!!!aaaaaaaaaaaa
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Answer:

8.85m/s

Explanation:

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When it strikes the ground, all of its Ep will transfer into Ek, so 1/2*m*v^2=78.4.

We already knew that m=2, so insert that in, we will get the V^2=78.4 m/s, V=8.85 m/s

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Is gravity a non-contact form?
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A group of students decides to set up an experiment in which they will measure the specific heat of a small amount of metal. The
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I think the answer is C
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4 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
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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

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Answer: Not 100% sure but I believe the answer is B.

Hope this helps! ^^
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