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:4.93 m/s
Explanation:
Given
height to reach is (h )1.24 m
here Let initial velocity is u
using equation of motion

here Final Velocity v=0
a=acceleration due to gravity




u=4.93 m/s
Distance traveled = 400 km
Volume of gas = 47.2 L
Note that
1 km = 0.62137 mi, or 1 mi = 1.609 km
1 L = 0.264 gal (US)
Therefore
400 km = (400 km)*(0.62137 mi/km) = 248.548 mi
47.2 L = (47.2 L)*(0.264 gal/L) = 12.4608 gal
The gas mileage is
(248.548 mi)/(12.4608 gal) = 19.95 mi/gal
Answer: 20.0 miles per gallon (nearest tenth)
Answer:
0.74 m/s
Explanation:
given,
time taken by the ramp to cover cliff , t = 64 s
acceleration = 0.37 m/s²
distance traveled by the belt

Clifford is moving with constant acceleration
...(1)
initial velocity is equal to zero
........(2)
equating equation (1) and (2)





Speed of the belt is equal to 0.74 m/s