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
6m at 0.9 radian means 6m at
Since positive angles are counter clockwise
6m at can be written as 6*cos
i+6*sin
j = 3.73i+4.70j
5m at can be written as 5*cos
i+5*sin
j = 1.294i-4.83j
4m at 1.2 radian means 4m at
4m at can be written as 4*cos
i+4*sin
j = 1.45i+3.73j
6m at can be written as 6*cos
i+6*sin
j = -5.20i-3j
a) So sum of the vector = 1.274i+0.6j
b) Magnitude =
c) Angle,
Answer:
Magnetic field, B = 0.004 mT
Explanation:
It is given that,
Charge,
Mass of charge particle,
Speed,
Acceleration,
We need to find the minimum magnetic field that would produce such an acceleration. So,
For minimum magnetic field,
B = 0.004 T
or
B = 4 mT
So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.