Answer:
q = 8.57 10⁻⁵ mC
Explanation:
For this exercise let's use Newton's second law
F = ma
where force is magnetic force
F = q v x B
the bold are vectors, if we write the module of this expression we have
F = qv B sin θ
as the particle moves perpendicular to the field, the angle is θ= 90º
F = q vB
the acceleration of the particle is centripetal
a = v² / r
we substitute
qvB = m v² / r
qBr = m v
q =
The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use
v = d / t
the distance is ¼ of the circle,
d =
d =
we substitute
v =
r =
let's calculate
r = 2 2.2 10-3 88 /πpi
r = 123.25 m
let's substitute the values
q = 7.2 10-8 88 / 0.6 123.25
q = 8.57 10⁻⁸ C
Let's reduce to mC
q = 8.57 10⁻⁸ C (10³ mC / 1C)
q = 8.57 10⁻⁵ mC
Answer:
58.37 V
Explanation:
Given that
Number of winding on coil, N = 80 turns
Area of the coil, A = 25 cm * 40 cm = 0.25 m * 0.4 m
Magnitude of magnetic force, B = 1.1 T
Time of rotation, t = 0.06 s
See the attachment for calculations
Well, if we've been paying attention in class, we already KNOW that the electrostatic force changes as the inverse square of the distance, and the top graph is conveniently labeled "Electrostatic Force".
But if we didn't already know that, we'd have to examine the graphs, and find the one where 'y' changes like 1/x² .
The top graph does that. After 1 unit of time, the force is 350. Double the time to 2 units, and the force should drop to 1/4 of 350 ... sure enough, it's a little less than 90. Double the time again, to 4 units, and it should drop to 1/4 of a little less than 90 ... by golly, it's down below 30.
The first graph is what an inverse square looks like. Now that you've worked out this graph, you'll know an inverse square relationship whenever you see it.
1,000 milligrams = 1 gram
2,000 milligrams = 2 grams
3,000 milligrams = 3 grams
4,000 milligrams = 4 grams
Answer:
(a) Velocity will be 17.146 m/sec
(b) Volume flow rate will be
Explanation:
We have given height of the dam h = 15 m
Acceleration due to gravity
Effective area
(a) From conservation of energy
Potential energy at the top will be equal to kinetic energy at the bottom
So
(b) We have to find the volume flow rate
Volume flow rate is given by