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:
instantaneous velocity is a velocity covered at an instant while average velocity is the change in distance/ the change in time taken
Answer:

Explanation:
<u>Instant Acceleration</u>
The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.
Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

And the acceleration is

Or equivalently

The given height of a projectile is

Let's compute the speed

And the acceleration

It's a constant value regardless of the time t, thus

Answer:
The kinetic energy of the particle as it moves through point B is 7.9 J.
Explanation:
The kinetic energy of the particle is:
<u>Where</u>:
K: is the kinetic energy
: is the potential energy
q: is the particle's charge = 0.8 mC
ΔV: is the electric potential = 1.5 kV
Now, the kinetic energy of the particle as it moves through point B is:


Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.
I hope it helps you!