We are given
V = 100 V
R = 500 Ω
C = 10^4 F
q(0) = 0
We are asked to find the charge q(t) of the RC capacitor.
To solve this, we use the formula:
q(t) = CV [1 - e^(-t/RC) ]
Substituting thet given values
q(t) = 10^4 (100) [1 - e^(-t/500(10^4) ]
q(t) = 1 x 20^6 [1 - e^(-2x10^7 t) ]
Since we only asked to get the charge q(t) in terms of t, the answer is
q(t) = 1 x 20^6 [1 - e^(-2x10^7 t) ]
Answer:
Explanation:
We use the following rotational kinematic equation to calculate the angular acceleration of the rod:
Here is the final angular speed, is the initial angular speed and is the angular acceleration. The rod starts rotating from rest ():
Recall that the angular speed is defined in function of the tangential speed (v) and the radius (r) of the circular motion:
In this case the radius is given by . Replacing (2) in (1):
Explanation:
Kinetic energy (J) = 2j
mass= 250g
velocity=?
1kg=1000g
mass= 250/1000
mass=0.25kg
Kinetic energy (J) = ½ x mass (kg) x [velocity]² (m/s)
2=1/2 × 0.25× [velocity]²
2=0.125× [velocity]²
[velocity]² = 2/0.125
[velocity]²=16
velocity= (16)^1/2
velocity= 4 m/s
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s