Answer:
t = 12,105.96 sec
Explanation:
Given data:
weight of spacecraft is 2000 kg
circular orbit distance to saturn = 180 km
specific impulse = 300 sec
saturn orbit around the sun R_2 = 1.43 *10^9 km
earth orbit around the sun R_1= 149.6 * 10^ 6 km
time required for the mission is given as t
![t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Csqrt%7B%5Cmu_sun%7D%7D%20%5B%5Cfrac%7B1%7D%7B2%7D%28R_1%20%2B%20R_2%29%5D%5E%7B3%2F2%7D)
where
is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2.![t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Csqrt%7B%201.32712%20x%2010%5E%7B20%7D%7D%7D%20%5B%5Cfrac%7B1%7D%7B2%7D%28149.6%20%2A%2010%5E%206%20%2B1.43%20%2A10%5E9%20%29%5D%5E%7B3%2F2%7D)
t = 12,105.96 sec
Answer:
π/10 rads
Explanation:
It takes an hour (60 minutes) for the minute's hand to turn a full circle or achieve an angular rotation of
2πl rad.
Now, number of periods of 3 minutes in an hour is;
Number of periods = 60/3 = 20 periods
Thus, 3 minutes rotation accounts for 1/20 of 2π the rotation of the minute's hand in an hour.
Thus;
Angular displacement = (1/20) * 2π = π/10 rads
Answer:
35%
Explanation:
The car's engine gives off 65% thermal energy
So only 35 % is converted into mechanical energy .
input heat = Q₁ = 100
output heat = Q₂ = 65
Work output = Q₁ - Q₂ = W
W = 100 - 65 = 35
Efficiency = W / Q₁ X 100
= (35/ 100) X 100
= 35%.
Answer:
Explanation:
The work done is defined as the product of force applied in the direction of displacement and the displacement.
W = F x d x Cosθ
where, F is the force applied, d be the displacement and θ be the angle between the displacement and force.
For the normal forces, the angle between the displacement and the force applied is 90 degree, and the value of Cos 90 is zero, so the work done is zero.
