I have read that it is enough energy to power the needs of civilization for an entire year.
We will use formula for the orbital velocity of Venus, which is v = 35.02 km/s.
An average distance to the Sun ( In kilometers ) is:
R = 0.723 * 149,579,871 km= 108,150,260 km.
Than we will calculate the orbital period ( T ).
v = 2 π R / T
T = 2 π R / v
T = 2 * 3.14 * 108,150,260 km / 126,072 km/s
T = 5389.75 s ≈ <span>224.5 days
The orbital period of Venus is approximately 224.5 days.</span>
Answer:his tension is 262 N
Explanation:1) Call T the tension of the rope2) Split T into its horizontal and vertical components.3) Horizontal component of the tension, Tx = T cos(55°)4) Vertical component of the tension, Ty = T sin (55°)5) Force equilibrium in the vertical direcction:∑Fy = 0Ty + normal force - weight of the sled = 0Call N the normal forceTy + N - 56 kg * 9.8 m/s^2Ty + N = 56 kg * 9.8 m/s^2Ty + N = 548.8NT sin(55) + N = 548.8N6) Force equilibrium in the horizontal directionconstant velocity => ∑Fx = 0Tx - Fx = 0Tcos(55) - Fx = 07) Fx is the friction force.The friction force and the normal force are related by the kinetic friction coefficient.Call μk the friction coefficientFx = μk N=> Tcos(55) - μk N = 0Tcos(55) - 0.45N = 08) Solve the system of two equations:Eq (1) T sin(55) + N = 548.8Eq (2) T cos(55) - 0.45N = 0Eq(1) 0.819T + N = 548.8Eq(2) 0.574T - 0.45N = 0The solution of the system is T = 262.01 N and N = 334.21 NThen T ≈ 262N
No work was done. Work is calculated by force/distance. If there isn't a distance change, no work has technically been done. I'm pretty sure that's right, at least! Hope it helps.
This is a job for Newton's 2nd Law! F = ma. We want to know the acceleration, so let's solve it for a, and get a = F/m. Now if we double the force, we'll have a' = 2F/m (where a' is the modified acceleration), and then if we decrease the mass by a half, we'd get a'' = 2F/(1/2*m) = 4F/m. We know the original acceleration was a = F/m, so let's sub that into the a'' equation and get: a'' = 4a, so a'' is 4 times the magnitude of our starting a.