Two passenger trains are passing each other on adjacent tracks. train a is moving east with a speed of 13 m/s, and train b is tr
2 answers:
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Okay, so using Kirchhoff's Loop Rule, we know the voltage drop across the entire system is 0. We can model this in a equation.
Vbatt = Vcap + Vres.
We are also told the voltage drop across the capacitor is I(t)R and voltage drop across the capacitor is q(t)/C. We can then substitute dq/dt in for I(t). We get:
Vbatt = q(t)/C + (dq/dt)R
This gets rid of the problem of I. Now we just have to get dq/dt by itself.
V - q/C = (dq/dt)*R Subtract q/C
(1/R)*(V - q/C) = dq/dt Divide by R on both sides
I think this is right, but don't quote me.
Hope this helps!
Vbatt = Vcap + Vres.
We are also told the voltage drop across the capacitor is I(t)R and voltage drop across the capacitor is q(t)/C. We can then substitute dq/dt in for I(t). We get:
Vbatt = q(t)/C + (dq/dt)R
This gets rid of the problem of I. Now we just have to get dq/dt by itself.
V - q/C = (dq/dt)*R Subtract q/C
(1/R)*(V - q/C) = dq/dt Divide by R on both sides
I think this is right, but don't quote me.
Hope this helps!
Answer:
The distance from Witless to Machmer is 438.63 m.
Explanation:
Given that,
Machmer Hall is 400 m North and 180 m West of Witless.
We need to calculate the distance
Using Pythagorean theorem
Where, =distance of Machmer Hall
=distance of Witless
Put the value into the formula
Hence, The distance from Witless to Machmer is 438.63 m.