Answer:
The time constant is 1.049.
Explanation:
Given that,
Charge 
We need to calculate the time constant
Using expression for charging in a RC circuit
![q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]](https://tex.z-dn.net/?f=q%28t%29%3Dq_%7B0%7D%5B1-e%5E%7B-%28%5Cdfrac%7Bt%7D%7BRC%7D%29%7D%5D)
Where,
= time constant
Put the value into the formula
![0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]](https://tex.z-dn.net/?f=0.65q_%7B0%7D%3Dq_%7B0%7D%5B1-e%5E%7B-%28%5Cdfrac%7Bt%7D%7BRC%7D%29%7D%5D)





Hence, The time constant is 1.049.
Answer: A)30V. First find the current of the circuit. I=V/R(total resistance). So I=60/120=0.5. Now to find voltage drop in R3 use ohms law as given. V(of 3)=(0.5)(60)=30V
Answer:
P_(pump) = 98,000 Pa
Explanation:
We are given;
h2 = 30m
h1 = 20m
Density; ρ = 1000 kg/m³
First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,
Thus, it can be expressed as;
P_(tank)+ P_(pump) = P_(nozzle)
Now, the pressure would be given by;
P = ρgh
So,
ρgh_1 + P_(pump) = ρgh_2
Thus,
P_(pump) = ρg(h_2 - h_1)
Plugging in the relevant values to obtain;
P_(pump) = 1000•9.8(30 - 20)
P_(pump) = 98,000 Pa
(a) The acceleration of the system is determined as 1.58 m/s².
(b) The relative weight of P is pounds is determined as 0.14 lb.
<h3>
Acceleration of the system</h3>
The acceleration of the system is calculated as follows;
W - T = m₂a --- (1)
T = m₁a ----(2)
μmgsinθ - m₁a = m₂a
(0.3 x 3 x 9.8 x sin40) - (0.4 + 0.2)a = 3a
5.67 - 0.6a = 3a
5.67 = 3.6a
a = 5.67/3.6
a = 1.58 m/s²
<h3>
Relative Weight of P</h3>
W = ma
W = 0.4 x 1.58
W = 0.632 N = 0.14 lb
Learn more about weight here: brainly.com/question/2337612
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