Answer:
23376 days
Explanation:
The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

Let the orbital period of the earth be
and its mean distance of from the sun be
.
Also let the orbital period of the planet be
and its mean distance from the sun be
.
Equation (2) therefore implies the following;

We make the period of the planet
the subject of formula as follows;

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

Substituting equation (5) into (4), we obtain the following;

cancels out and we are left with the following;

Recall that the orbital period of the earth is about 365.25 days, hence;

Answer:
A transformed, visible pattern on earth that holds cues indicating the presence of a particular ethnic group is cultural landscape.
Explanation:
Answer:
Explanation:
Mass of ice m = 500g = .5 kg
Heat required to raise the temperature of ice by 10 degree
= mass of ice x specific heat of ice x change in temperature
= .5 x 2093 x 10 J
10465 J
Heat required to melt the ice
= mass of ice x latent heat
0.5 x 334 x 10³ J
167000 J
Heat required to raise its temperature to 18 degree
= mass x specific heat of water x rise in temperature
= .5 x 4182 x 18
=37638 J
Total heat
=10465 +167000+ 37638
=215103 J
Answer: Search Results
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Answer: Surface waves can have characteristics of both longitudinal and transverse waves in the following way; The motion of the surface waves is up and down which is perpendicular to the direction of the wave. This is similar to the motion of transverse waves whereas the the motion of longitudinal.
Explanation:
Answer:
the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.
Explanation:
Mass of satellite, m
orbit radius of first, r1 = r
orbit radius of second, r2 = 2r
Centripetal force is given by

Where v be the orbital velocity, which is given by

So, the centripetal force is given by

where, g bet the acceleration due to gravity

So, the centripetal force

Gravitational force on the satellite having larger orbit
.... (1)
Gravitational force on the satellite having smaller orbit
.... (2)
Comparing (1) and (2),
F' = 4 F
So, the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.