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
Answer:
12.50 m/s
Explanation:
Vi = 9.49 m/s
a = 0.988 m/s²
t = 3.05 s
Vf = ?
Vf = Vi + at
Vf = 9.49 + (0.988)(3.05)
Vf = 12.50 m/s
A). 1,000 watts = 1 kilowatt
5,000 watts = 5 kilowatts
b). (5 kilowatts) x (2 hours) = 10 kilowatt-hours (kWh)
c). (10 kWh) x (15 cents/kWh) = $1.50
Answer:
The number density of the gas in container A is twice the number density of the gas in container B.
Explanation:
Here we have
P·V =n·R·T
n = P·V/(RT)
Therefore since V₁ = V₂ and T₁ = T₂
n₁ = P₁V₁/(RT₁)
n₂ = P₂V₂/(RT₂)
P₁ = 4 atm
P₂ = 2 atm
n₁ = 4V₁/(RT₁)
n₂ =2·V₁/(RT₁)
∴ n₁ = 2 × n₂
Therefore, the number of moles in container A is two times that in container B and the number density of the gas in container A is two times the number density in container B.
This can be shown based on the fact that the pressure of the container is due to the collision of the gas molecules on the walls of the container, with a kinetic energy that is dependent on temperature and mass, and since the temperature is constant, then the mass of container B is twice that of A and therefore, the number density of container A is twice that of B.
