Answer:
14 hours 18 minutes.
Explanation:
ratio of number of orbits, so it completes 7 orbits in the time Janus does 6.
(16*60+41)*6/7=858 minutes or 14 hours 18 minutes
Answer:
Power = 124.50 W
Explanation:
Given that:
The Sound intensity of a speaker output is 102 dB
and the distance r = 25 m
For the intensity of sound,

where;
the threshold of hearing 



I = 0.01585 W/m²
If we recall, we know remember that ;
Power = Intensity × A
rea
Power = 0.01585 W/m² × 4 × 3.142 × (25 m)²
Power = 124.50 W
Answer:
0.00903 rad
0.00926 rad
6.268\times 10^{-6}
Explanation:
s = Diameter of the object
r = Distance between the Earth and the object
Angle subtended is given by

For the Moon

The angle subtended by the Moon is 0.00903 rad
For the Sun

The angle subtended by the Sun is 0.00926 rad
Area ratio is given by

The area ratio is 
Complete question :
NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?
Answer:
601000 N
Explanation:
Given that :
Acceleration due to gravity at lunar outpost = 1.6m/s²
Supported Weight of supplies = 1 * 10^5 N
Acceleration due to gravity on the earth surface = 9.8m/s²
Maximum weight of supplies as measured on EARTH :
Ratio of earth gravity to lunar post gravity:
(Earth gravity / Lunar post gravity) ;
(9.8 / 1.63) = 6.01
Hence, maximum weight of supplies as measured on EARTH should be :
6.01 * (1 × 10^5)
6.01 × 10^5
= 601000 N