Answer:
213 nA
2.13 mA
851e^-t μA
Explanation:
We have a pretty straightforward question here.
Ohms Law states that the current in an electric circuit is directly proportional to the voltage and inversely proportional to the resistance in the circuit. It is mathematically written as
V = IR, since we need I, we can write that
I = V/R
a) at V = 1 mV
I = (1 * 10^-3) / 4.7 * 10^3
I = 2.13 * 10^-7 A or 213 nA
b) at V = 10 V
I = 10 / 4.7 * 10^3
I = 0.00213 A or 2.13 mA
c) at V = 4e^-t
I = 4e^-t / 4.7 * 10^3
I = 0.000851e^-t A or 851e^-t μA
Answer:
The impulse on the object is 60Ns.
Explanation:
Impulse is defined as the product of the force applied on an object and the time at which it acts. It is also the change in the momentum of a body.
F = m a
F = m(
)
⇒ Ft = m(
-
)
where: F is the dorce on the object, t is the time at which it acts, m is the mass of the object,
is its initialvelocity and
is the final velocity of the object.
Therefore,
impulse = Ft = m(
-
)
From the question, m = 3kg,
= 0m/s and
= 20m/s.
So that,
Impulse = 3 (20 - 0)
= 3(20)
= 60Ns
The impulse on the object is 60Ns.
Answer:
I = Δq / t
Explanation:
The quantity of electricity i.e charge is related to current and time according to the equation equation:
Q = It
Δq = It
Where:
Q => is the quantity of electricity i.e charge
I => is the current.
t => is the time.
Thus, we can rearrange the above expression to make 'I' the subject. This is illustrated below:
Δq = It
Divide both side by t
I = Δq / t
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