D. Velocity because it describes a speed and direction
The area of the Earth (Ae) that is irradiated by is given by:
Ae = 4πRe^2, where Re = Distance from Sun to Earth
Substituting;
Ae = 4π*(1.5*10^8*1000)^2 = 2.827*10^23 m^2
On the Earth, insolation (We) = Psun/Ae
Therefore,
Psun (Rate at which sun emits energy) = We*Ae = 1.4*2.827*10^23 = 3.958*10^23 kW = 3.958*10^26 W
The crate would slide forward
We apply the following equation
T = 2π * sqrt (L/g)
Where g is the gravity = 9.8 m/s^2
L is the longitude of the pendulum (Height of the tower)
T is the period. (T = 18s)
We find L.............> (T /2π)^2 = L/g
L = g*(T /2π)^2...........> L = 80.428 meters
Answer:
The final speed of the stone as it lift the ground is 23.86 m/s.
Explanation:
Given that,
Force acting on the rock, F = 3 N
Distance, d = 16 m
Initial speed of the stone, u = 22 m/s
We need to find the rock's speed just as it left the ground. It can be calculated using work energy theorem as :
![W=\Delta E\\\\W=\dfrac{1}{2}m(v^2-u^2)\\\\Fd=\dfrac{1}{2}m(v^2-u^2)\\\\v^2=\dfrac{2Fd}{m}+u^2\\\\v^2=\dfrac{2mgd}{m}+u^2\\\\v^2=2\times 9.8\times 16+(16)^2\\\\v=23.86\ m/s](https://tex.z-dn.net/?f=W%3D%5CDelta%20E%5C%5C%5C%5CW%3D%5Cdfrac%7B1%7D%7B2%7Dm%28v%5E2-u%5E2%29%5C%5C%5C%5CFd%3D%5Cdfrac%7B1%7D%7B2%7Dm%28v%5E2-u%5E2%29%5C%5C%5C%5Cv%5E2%3D%5Cdfrac%7B2Fd%7D%7Bm%7D%2Bu%5E2%5C%5C%5C%5Cv%5E2%3D%5Cdfrac%7B2mgd%7D%7Bm%7D%2Bu%5E2%5C%5C%5C%5Cv%5E2%3D2%5Ctimes%209.8%5Ctimes%2016%2B%2816%29%5E2%5C%5C%5C%5Cv%3D23.86%5C%20m%2Fs)
So, the final speed of the stone as it lift the ground is 23.86 m/s.