Answer:
a) The speed is 61.42 m/s
b) The drag force is 10.32 N
Explanation:
a) The Reynold´s number for the model and prototype is:


Equaling both Reynold's number:

Clearing Vm:

b) The drag force is:

I think there's a typo because the answer I'm getting is very large.
This is what I'm getting
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c = speed of light
c = 3.0 x 10^8 m/sec approximately
This is roughly 300 million meters per second
The time it takes the signal to reach the aircraft and come back is 1.4 x 10^3 seconds. Half of this time period is going one direction (say from the radar station to the aircraft), so (1.4 x 10^3)/2 = 7.0 x 10^2 seconds is spent going in this one direction.
distance = rate*time
d = r*t
d = (3.0 x 10^8) * (7.0 x 10^2)
d = (3.0*7.0) x (10^8*10^2)
d = 21.0 x 10^(8+2)
d = 21.0 x 10^10
d = (2.1 x 10^1) * 10^10
d = 2.1 x (10^1*10^10)
d = 2.1 x 10^11 meters
d = 210,000,000,000 meters (this is 210 billion meters; equivalent to roughly 130,487,950 miles)
Answer:
The resulting magnetic force on the wire is -1.2kN
Explanation:
The magnetic force on a current carrying wire of length 'L' with current 'I' in a magnetic field B is
F = I (L*B)
Finding (L * B) , where L = (2, 0, 0)m , B = (30, -40, 0)
L x B =
= (0, 0, -80)
we can now solve
F = I (L x B) = I (-80)
F = -1200 kmN
F = -1200 kN * 10⁻³
F = -1.2kN
Answer:
14 m/s
Explanation:
Using the principle of conservation of energy, the potential energy is converted to kinetic energy, assuming any losses.
Kinetic energy is given by ½mv²
Potential energy is given by mgh
Where m is the mass, v is the velocity, g is acceleration due to gravity and h is the height.
Equating kinetic energy to be equal to potential energy then
½mv²=mgh
V
Making v the subject of the formula
v=√(2gh)
Substituting 9.81 m/s² for g and 10 m for h then
v=√(2*9.81*10)=14.0071410359145 m/s
Rounding off, v is approximately 14 m/s