Answer:
Explanation:
Just start with the trivial. If gravity was a constant then E = mgh and 1/2 m v^2 = E
so v=sqrt(2gh)=sqrt(2*9.8*1590000) = 5582m/s
Now this will be too high as gravity reduces with distance.
However it is still true that 1/2mv^2 = loss of gravitational potential energy
so 1/2 v^2 = loss of gravitational potential ( i.e a field without considering mass )
As g = GM/ Ro^2 and P = - GM/R
the Po = - 9.8 * (6370*10^3)= - 62.4 * 10 ^ 6 J/kg
P1= Po * 6370/(6370+1590) = - 49.93 * 10 ^ 6 J/kg
find the CHANGE and then from that the velocity
ie v = sqrt(2*( P1 - Po)) = 5094 m/s
Note how it is a bit smaller than the first estimate but not by such a margin that they are unrecognizably different.
there here is your answer to your question
Uncertainty in measurements and calculations means difference between actual and measured data. We can say that all measurements have some degree of uncertainty. ... Systematic error (because of error in measuring instrument) 2. Random error (human errors such as- delay in starting, delay in stopping).
Light is either potential energy or kinetic.
Answer: 71.7 KJ
Explanation:
The rotational kinetic energy of a rotating body can be written as follows:
Krot = ½ I ω2
Now, any point on the rim of the flywheel, is acted by a centripetal force, according to Newton’s 2nd Law, as follows:
Fc = m. ac
It can be showed that the centripetal acceleration, is related with the angular velocity and the radius, as follows:
ac = ω2 r
We know that this acceleration has a limit value, so , we can take this limit to obtain a maximum value for the angular velocity also.
As the flywheel is a solid disk, the rotational inertia I is just ½ m r2.
Replacing in the expression for the Krot, we have:
Krot= ½ (1/2 mr2.ac/r) = ¼ mr ac = ¼ 67.0 Kg. 1.22 m . 3,510 m/s2 = 71. 7 KJ