Answer:Orbital period =21.22hrs
Explanation:
given that
mass of earth M = 5.97 x 10^24 kg
radius of a satellite's orbit, R= earth's radius + height of the satellite
6.38X 10^6 + 3.25 X10^7 m =3.89 X 10^7m
Speed of satellite, v= 
where G = 6.673 x 10-11 N m2/kg2
V= \sqrt (6.673x10^-11 x 5.97x10^ 24)/(3.89 X 10^ 7m)
V =10,241082.2
v= 3,200.2m/s
a) Orbital period
= 
V=
T= 2 
r/ V
= 2 X 3.142 X 3.89 X 10^7m/ 3,200.2m/s
=76,385.1 s
60 sec= 1min
60mins = 1hr
76,385.1s =hr
76,385.1/3600=21.22hrs
Bit of an odd question. Power Plants are known to use water-powered turbines to generate electricity, but can also make use of nuclear fission.
Answer:
by a factor of 2
Explanation:
Maximum speed of a body in simple harmonic motion relate to the amplitude by the following formula:
v ( maximum speed in m/s ) = x ( amplitude in meters ) √K /m where K is in N/m and m is kg
v is directly proportional to the amplitude and increases as the amplitude increases by a factor of 2
Answer:
10 s
Explanation:
Given:
Δx = 100 m
v₀ = 0 m/s
a = 2 m/s²
Find: t
Δx = v₀ t + ½ at²
100 m = (0 m/s) t + ½ (2 m/s²) t²
t = 10 s
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