Answer:
The magnitude of the magnetic force at 32.5⁰ to the field, is 2.22 F
Explanation:
Given;
magnitude of the magnetic force is F at 14.0⁰ with respect to the field.
To determine the magnitude of the magnetic force at 32.5⁰ to the field, we apply the following formula and solve with proportion;
f*Sin(14.0⁰) = F
f*Sin(32.5⁰) = ?
Therefore, the magnitude of the magnetic force at 32.5⁰ to the field, is 2.22 F
Answer:
Minimum capacitance = 200 μF
Explanation:
From image B attached, we can calculate the current flowing through the capacitors.
Thus;
Since V=IR; I = V/R = 5/500 = 0.01 A
Maximum charge in voltage is from 5V to 4.9V. Thus, each capacitor will have 2.5V. Hence, change in voltage(Δv) for each capacitor will be ; Δv = 0.05 V
So minimum capacitance will be determined from;
i(t) = C(dv/dt)
So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]
C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF
Given:
Mass of the rail road car, m = 2 kg
velocity of the three cars coupled system, v' = 1.20 m/s
velocity of first car, = 3 m/s
Solution:
a) Momentum of a body of mass 'm' and velocity 'v' is given by:
p = mv
Now for the coupled system according to law of conservation of momentum, total momentum of a system before and after collision remain conserved:
(1)
where,
= velocity of the first car
= velocity of the 2 coupled cars after collision
Now, from eqn (1)
v' = 1.80 m/s
Therefore, the velocity of the combined car system after collision is 1.80 m/s