Explanation:
The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :
We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :
At t = 1 s,
Current,
So, the current at t = 1 s is 3 A.
For lowest current,
Hence, this is the required solution.
To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m
The potential difference is related to the electric field by:
∆V=Ed
where,
∆V is the potential difference
E is the electric field
d is the distance
what is potential difference?
The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.
We want to know the distance the detectors have to be placed in order to achieve an electric field of
E=1v/cm=100v/cm
when connected to a battery with potential difference
∆v=1.5v
Solving the equation,we find
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The velocity at point B is 10 m/s with a pressure of 262500 Pa
<h3>Bernoulli equation</h3>
According to the continuity equation:
A₁V₁ = A₂V₂
Where A is the area and V is velocity, ρ = density of water = 1000 kg/m³
Hence:
4(5) = 2(V₂)
V₂ = 10 m/s
Using Bernoulli equation:
The velocity at point B is 10 m/s with a pressure of 262500 Pa
Find out more on Bernoulli equation at: brainly.com/question/14082066