Answer:
Orbital speed=8102.39m/s
Time period=2935.98seconds
Explanation:
For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)
V2R+h=g(R2(R+h)2)
V=√g(R2R+h)
V= sqrt(9.8 × (6371000)^2/(6371000+360000)
V= sqrt(9.8× (4.059×10^13/6731000)
V=sqrt(65648789.18)
V= 8102.39m/s
Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)
T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)
T=sqrt(3.40×10^21)/ (3.99×10^14)
T= sqrt(0.862×10^7)
T= 2935.98seconds
In general,
Power = (energy moved) / (time to move the energy) .
If it's mechanical power, then
Power = (work done) / (time to do the work) .
If it's electrical power, then it can be any one of these:
Power = (volts) x (amperes)
Power = (volts)² / (resistance, ohms)
Power = (amperes)² x (resistance, ohms) .
Whatever kind of energy you're dealing with, power always
turns out to be
(amount of energy produced, used, or moved)
divided by
(time taken to produce, use, or move the energy) .
Answer:
22.505 seconds
Explanation:
V =19.8m/s
V = a*to
t1 = 19.8/3.3
= 6seconds
Distance travelled during acceleration
= 1/2 x 3.3 x 6²
= 59.4m
X_total = x1 + x2
X2 = 373-59.4
X2 = 313.6m
t2 = x2/v
= 313.6/19.8
= 16.505
Total = 16.505 + 6
= 22.505 seconds
the minimum time in which an elevator can travel the 373 m from the ground floor is 22.505 seconds.
The pressure at the depth 11 km below sea level can be
calculated using
P=ρgh
P is pressure, ρ is the density of the fluid; g is the
gravitational constant, h is the height from the surface, or depth that the
object is submerged.
P = ( 1000 kg/ m3) ( 9.81 m.s2)( 11 000m) + 1 atm
P = 107,910,000 pa ( 1 atm/ 101 325 Pa) + 1 atm = 1066 atm