The relevant equation we can use in this problem is:
h = v0 t + 0.5 g t^2
where h is height, v0 is initial velocity, t is time, g is
gravity
Since it was stated that the rock was drop, so it was free
fall and v0 = 0, therefore:
h = 0 + 0.5 * 9.81 m/s^2 * (4.9 s)^2
<span>h = 117.77 m</span>
Answer:
Orbital speed=8102.39m/s
Time period=2935.98seconds
Explanation:
For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)
V2R+h=g(R2(R+h)2)
V=√g(R2R+h)
V= sqrt(9.8 × (6371000)^2/(6371000+360000)
V= sqrt(9.8× (4.059×10^13/6731000)
V=sqrt(65648789.18)
V= 8102.39m/s
Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)
T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)
T=sqrt(3.40×10^21)/ (3.99×10^14)
T= sqrt(0.862×10^7)
T= 2935.98seconds
Answer:
Anybody which is in state of rest ,will be in rest if we don't apply any external force ...
Answer:
The self-induced emf in this inductor is 4.68 mV.
Explanation:
The emf in the inductor is given by:
Where:
dI/dt: is the decreasing current's rate change = -18.0 mA/s (the minus sign is because the current is decreasing)
L: is the inductance = 0.260 H
So, the emf is:
Therefore, the self-induced emf in this inductor is 4.68 mV.
I hope it helps you!
