Answer:
r = 1.63×10^5 mi
Explanation:
Let r = distance of object from earth
Rs = distance between earth and sun
Ms = mass of the sun
= 3.24×10^5 Me (Me = mass of earth)
At a distance R from earth, the force Fs exerted by the sun on the object is equal to the force Fe exerted by the earth on the object. Using Newton's universal law of gravitation,
Fs = Fe
GmMs/(Rs - r)^2 = GmMe/r^2
This simplifies to
Ms/(Rs - r)^2 = Me/r^2
(3.24×10^5 Me)/(Rs - r)^2 = Me/r^2
Taking the reciprocal and then its square root, this simplifies further to
Rs - r = (569.2)r ----> Rs = 570.2r
or
r = Rs/570.2 = (9.3×10^7 mi)/570.2
= 1.63×10^5 mi
Answer:
Time, t = 2 seconds
Explanation:
Given the following data;
Mass, m = 50 kg
Initial velocity, u = 0 m/s (since it's starting from rest).
Final velocity, v = 8 m/s
Force, F = 200 N
To find the time, we would use the following formula;
Making time, t the subject of formula, we have;
Substituting into the formula, we have;
Time, t = 2 seconds
Power = Work done / Time taken.
Work done = mgh
Mass, m = 33kg ( Am presuming it is 33 kg).
h = 85 m.
Work done = 33 * 9.81* 85 = 27517.05 J.
Time taken.
Since object was dropped from height, it fell under gravity.
Using H = ut + (1/2) * gt^2. u = 0.
H = 1/2 gt^2.
t = (2H/g) ^ (1/2)
t = (2*85/9.81) ^ 0.5 = 4.1628 s.
Power = 27517.05 / 4.1628 = 6610.23 Watts.
= 6610 W to 3 S. f.
Location & Sunlight Availability.
Solar Panels use a large amount of space.
The Sun isn't always present.
Solar Energy is Inefficient.
There is an overlooked Pollution & Environmental Impact.
Expensive Energy Storage.
High Initial Cost.
The answer would be to research the need. This should have been done before the project began.
