The solution would be like
this for this specific problem:
<span>5.5 g = g + v^2/r </span><span>
<span>4.5 g =
v^2/r </span>
<span>v^2 = 4.5
g * r </span>
<span>v = sqrt
( 4.5 *9.81m/s^2 * 350 m) </span>
v = 124
m/s</span>
So the pilot will black out for this dive at 124
m/s. I am hoping that these answers have satisfied your query and it
will be able to help you in your endeavors, and if you would like, feel free to
ask another question.
At each point on a 'line', the direction of the 'line' is the direction of the force
on a small test magnet placed in the field at that point.
If two 'lines' crossed at the same point, that means a small test magnet placed
at that point in the field would feel a force in two different directions.
But even if that were true, then the net effect on the small test magnet would be
the vector sum of the two forces, and they would be represented by a single net
force anyway, and therefore by a single field 'line' at that point.
Answer:
80
Explanation:
because of the denstiy formula
4395g = 4.395 . divide it by 1000
Answer:
1. e m fmax = 0.00598 Volt
2. Imax = 0.000854 Amp
Explanation:
1. Find the maximum induced emf.
e m fmax =
Given that e m fmax = N*A*B*w
N = 1
A = 2 cm^2 = 0.0002 m^2
f = 143 rotation per minute = 143/min
f = (143/min) * (1 min/60 sec) = 2.38/sec
w = 2Πf = 2 * Π * 2.38 = 14.95 rad/sec
B = 2T
e m fmax = N*A*B*w
e m fmax = 1 * 0.0002 * 2 * 14.95
e m fmax = 0.00598 Volt.
2. Find the maximum current through the bulb.
Imax = e m fmax / R
Where R is the total resistance in the circuit is 7 Ω.
Imax = 0.00598/7 = 0.000854 Amp.
Imax = 0.000854 Amp