Answer: 200 knots
Explanation: the maximum indicated airspeed at which aircraft may be flown when at or below 2,500 feet AGL and within 4 nautical miles of the primary airport of Class C airspace is 200 KNOTS
<span>EP (potential energy) = mgy -> (59)(9.8)(-5) = -2,891
EP + EK (kinetic energy) = 0; but rearranging it for EK makes it EK = -EP, such that EK = 2891 when plugged in.
EK = 0.5mv^2, but can also be v = sqrt(2EK/m).
Plugging that in for sqrt((2 * 2891)/59), we get 9.9 m/s^2 with respect to significant figures.</span>

According to above question ~
Let's find the charge (q) by using formula ~
Hence, 12 coulombs of charge flow past any point in the wire in 3 seconds
Answer:
The friction force acting on the object is 7.84 N
Explanation:
Given;
mass of object, m = 4 kg
coefficient of kinetic friction, μk = 0.2
The friction force acting on the object is calculated as;
F = μkN
F = μkmg
where;
F is the frictional force
m is the mass of the object
g is the acceleration due to gravity
F = 0.2 x 4 x 9.8
F = 7.84 N
Therefore, the friction force acting on the object is 7.84 N
Given parameters:
Displacement = 8km
Velocity = 3.8km/h
Unknown:
time = ?
Solution:
Velocity is displacement divided by time.
Velocity =
Displacement = velocity x time
Input the parameters:
8 = 3.8 x time
Time =
= 2.1s
The time taken is 2.1s