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
The velocity of the second glider after the collision is 4.33 m/s rightward.
<h3>
Velocity of the second glider after the collision</h3>
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where;
- m₁ is mass of first glider
- m₂ is mass of second glider
- u₁ is initial velocity of first glider
- u₂ is initial velocity of second glider
- v is the final velocity of the gliders
(2)(1) + (3)(5) = (2)(2) + 3v₂
17 = 4 + 3v₂
3v₂ = 17 - 4
3v₂ = 13
v₂ = 13/3
v₂ = 4.33 m/s
Thus, the velocity of the second glider after the collision is 4.33 m/s rightward.
Learn more about linear momentum here: brainly.com/question/7538238
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