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
Answer:
The correct relationships are T-fg=ma and L-fg=0.
(A) and (C) is correct option.
Explanation:
Given that,
Weight Fg = mg
Acceleration = a
Tension = T
Drag force = Fa
Vertical force = L
We need to find the correct relationships
Using balance equation
In horizontally,
The acceleration is a
...(I)
In vertically,
No acceleration
Put the value of mg
....(II)
Hence, The correct relationships are T-fg=ma and L-fg=0.
(A) and (C) is correct option.
Answer:
1 angstrom = 0.1nm
5000 angstrom = 5000/1 × 0.1nm
<h3>= 500nm</h3>5000 angstrom = 5000 × 1 × 10^-10
<h3>= 5 × 10^-7 m</h3>Hope this helps you