Consider the equation . The dimensions of the variables v, x, and t are , , and , respectively. The numerical factor 3 is dimens
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Answer:
Explanation:
As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from to
is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.
The total distance covered by the train during the entire journey is the area of the speed-time graph.
Area
As velocity is in and time is in
so the unit of area is
Hence, the total distance is .
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