Answer:
a) a geostationary satellite is that it is always at the same point with respect to the planet,
b) f = 2.7777 10⁻⁵ Hz
c) d) w = 1.745 10⁻⁴ rad / s
Explanation:
a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet
- T = 10 h (3600 s / 1h) = 3.6 104 s
b) the period the frequency are related
T = 1 / f
f = 1 / T
f = 1 / 3.6 104
f = 2.7777 10⁻⁵ Hz
c) the distance traveled by the satellite in 1 day
The distance traveled is equal to the length of the circumference
d = 2pi (R + r)
d = 2pi (69 911 103 + 120 106)
d = 1193.24 m
d) the angular velocity is the angle traveled between the time used.
.w = 2pi /t
w = 2pi / 3.6 10⁴
w = 1.745 10⁻⁴ rad / s
how fast is
v = w r
v = 1.75 10-4 (69.911 106 + 120 106)
v = 190017 m / s
Answer:
A) Average speed = 18.75 m/s
B) More time is spent at 15 m/s than at 25 m/s.
Explanation:
Let the first distance be d1 and the second distance be d2.
We are given;
d1 = 10 km = 10000 m
d2 = 10 km = 10000 m
Speed; v1 = 15 m/s
Speed; v2 = 25 m/s
Now, the formula for distance is; Distance = speed x time
Thus:
d1 = v1 x t1
t1 = d1/v1 = 10000/15 = 666.67 seconds
Also,
d2 = v2 x t2
t2 = d2/v2 = 10000/25 = 400 seconds
Average speed = total distance/total time = (10000 + 10000)/(666.67 + 400) = 18.75 m/s
From earlier, since t1 = 666.67 seconds and t2 = 400 seconds, then;
More time at 15 m/s than at 25 m/s.
Answer:
49N
Explanation:
F=ma
We know the mass is 5kg, and since the ball is suspended on one cable, the acceleration is g, 9.8m/s^2
F=5kg*9.8m/s^2
= 49N
Hope this helps!
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 
.
