Atom* the particles are (Electrons)
Answer:
I will specify a value of 0.009T for the alternator’s magnetic field
Explanation:
E_peak = 14 V
d = 10cm = 0.1m, so r = 0.1/2 =0.05m
N = 250 turns
f = 1200rpm = (1200rp/m x 1m/60sec) = 20 revolutions per second
At peak performance, peak voltage is given by the equation;
E_peak = NABω
Let's make the magnetic field B the subject;
B = E_peak/(NAω)
Now we know that ω = 2πf
Thus, ω = 2π x 20 revs/s = 125.664 revs/s.
Let's convert it to the standard unit which is rad/s.
1 rev/s = 6.283 rad/s
Thus, 125.664 revs/s = 125.664 x 6.283 = 789.55 rad/s
Area (A) = πr² = π x 0.05² = 0.007854 m²
Thus, plugging in the relevant values to get;
B = 14/[(250 x 0.007854 x 789.55)] = 0.009T
The net force on an object is:
F = ma
The only force acting on the speck of dust as it is lays on the CD is centripetal force given by:
F = (mv²)/r
Equating the two
ma = (mv²)/r
We get:
a = v²/r
v is the linear velocity in this case; however, we can calculate only the angular velocity with the given data. Therefore, we must use:
v = ωr; where
ω = 2πf; f is the rotations per second
f = 9500 / 60
f = 950/6
ω = 2π(950/6)
ω = (950π)/3
v = (950π)/3 × 0.12
v = 38π
a = (38π)²/0.12
a = 1.2 × 10⁵ m/s²
To express this in terms of gravitational acceleration, we divide by the value of gravitational acceleration, 9.81
a = 1.2 × 10⁴ g
The force of gravity between two objects is given by:
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation
In this problem, the mass of the object is
, while the Earth's mass is
. Their separation is
, therefore the gravitational force exerted on the object is
