DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t) by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150
Answer:
I think is Number 2
The rate at which calcium chloride is equal to the rate at which sodium chloride is produced.
Let me know if I'm wrong.
Answer:
'A' is the the point on the graph that shows a temperature of 40°C and the time of 25 minutes
Answer:
Magnetic field, B = 0.275 T
Explanation:
Given that,
Length of the wire, L = 35 cm = 0.35 m
Current carried in the wire, I = 2.6 A
The segment makes an angle of 53∘ with the direction of the magnetic field, 
Magnetic force, F = 0.2 N
To find,
The magnitude of the magnetic field.
Solution,
The magnetic force acting on the wire is given by :

is the angle between the length of wire and the magnetic field.

B = 0.275 T
Therefore, the magnitude of the magnetic field is 0.275 T. Hence, this is the required solution.