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
Complete Question
The speed of a transverse wave on a string of length L and mass m under T is given by the formula
If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
Answer:
Explanation:
From the question we are told that
Speed of a transverse wave given by
Maximum Tension is 
Generally making 
subject from the equation mathematically we have
Therefore the Linear mass in terms of Velocity is given by
Answer:
2 charges of electron (2C)
Explanation:
I = Q/t
2 = Q/1
Q = 2×1= 2C
Q = 2 charge of electron
Given:-
- Time taken by the particle (t) = 6 s
- Average speed (v) = 40 m/s
To Find: Distance (s) travelled by the particle.
We know,
s = vt
where,
- s = Distance travelled,
- v = Speed &
- t = Time taken.
Putting the values,
s = (40 m/s)(6 s)
→ s = 240 m ...(Ans.)
