Answer:
7.65x10^3 m/s
Explanation:
The computation of the satellite's orbital speed is shown below:
Given that
Earth mass, M_e = 5.97 × 10^24 kg
Gravitational constant, G = 6.67 × 10^-11 N·m^2/kg
Orbital radius, r = 6.80 × 10^6m
Based on the above information
the satellite's orbital speed is
V_o = √GM_e ÷ √r
= √6.67 × 10^-11 × 5.97 × 10^24 ÷ √6.80 × 10^6
= 7.65x10^3 m/s
Answer:
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
b) Determine the magnitude and direction of current in wire 3
Explanation:
a) 

position of wire = 50 - 1.2
= 48.8cm
b) 

Direction ⇒ downward
The particles vibrate against each other in a soild state........i think