It depends on both of them.
In fact, the projectile begins its motion with an initial velocity of
![v_0](https://tex.z-dn.net/?f=v_0)
and an angle of
![\alpha](https://tex.z-dn.net/?f=%5Calpha)
. On the y-axis (vertical direction), it is an accelerated motion with acceleration equal to -g (gravitational acceleration). The vertical velocity of the projectile at any time t is given by
![v_y (t) = v_0 sin \alpha + gt](https://tex.z-dn.net/?f=v_y%20%28t%29%20%3D%20v_0%20sin%20%5Calpha%20%2B%20gt%20)
and as it can be seen, this depends on both initial velocity and angle.
Answer:
6.88 mA
Explanation:
Given:
Resistance, R = 594 Ω
Capacitance = 1.3 μF
emf, V = 6.53 V
Time, t = 1 time constant
Now,
The initial current, I₀ = ![\frac{\textup{V}}{\textup{R}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctextup%7BV%7D%7D%7B%5Ctextup%7BR%7D%7D)
or
I₀ = ![\frac{\textup{6.53}}{\textup{594}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctextup%7B6.53%7D%7D%7B%5Ctextup%7B594%7D%7D)
or
I₀ = 0.0109 A
also,
I = ![I_0[1-e^{-\frac{t}{\tau}}]](https://tex.z-dn.net/?f=I_0%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%7D%5D)
here,
τ = time constant
e = 2.717
on substituting the respective values, we get
I = ![0.0109[1-e^{-\frac{\tau}{\tau}}]](https://tex.z-dn.net/?f=0.0109%5B1-e%5E%7B-%5Cfrac%7B%5Ctau%7D%7B%5Ctau%7D%7D%5D)
or
I =
or
I = 0.00688 A
or
I = 6.88 mA
Just treat the bead as a point charge.
E = kq/r^2, and E points away from the center of the sphere since the charge is positive.
True. E=energy M=mass c= speed of light