Crescent, gibbous, waxing, and waning.
To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:

Where,
G = Gravitational Universal Constant
M = Mass of Planet
r = Radius of the planet ('h' would be the orbit from the surface)
The escape velocity is

Through this equation we can find the mass of the Planet in function of the distance, therefore



The orbital velocity is





The time period of revolution is,




Therefore the orbital period of the satellite is closes to 1 hour and 12 min
Answer:
0.000025s
Explanation:
Period it’s. : T(s)= 1/f(Hz)=1/40000Hz=0.000025s
Answer:
B meet A 0.01 km east of flagpole
Explanation:
given data
distance A = 5.7 km west
velocity V1 = 8.9 km/h
distance B = 4.5 km east
velocity V2 = 7 km/h
to find out
How far runners from the flagpole, when paths cross
solution
we know A and B are 5.7 + 4.5 = 10.2 km apart
and we consider here B will run distance x km for meet
so time will be for B is
time B = distance / velocity
time B = x / 7 ...................1
and
for A distance for meet = ( 10.2 - x ) km
so time A = distance / velocity
time A = ( 10.2 - x ) / 8.9 .............2
now equating equation 1 and 2
time A = time B
x / 7 = ( 10.2 - x ) / 8.9
x = 4.490
so distance of B run for meet is 4.490 km
so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km
so B meet A 0.01 km east of flagpole
Answer:
Explanation:
according to ohms law we know that
v=IR
given current =2 amps
given resistance =6Ω
so voltage is
v=2*6 =12 V