<span>1) The differential equation that models the RC circuit is :
(d/dt)V_capacitor </span>+ (V_capacitor/RC) = (V_source/<span>RC)</span>
<span>Where the time constant of the circuit is defined by the product of R*C
Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s
2) C<span>harge of the capacitor 1.57 time constants
1.57*(2.742) = 4.3048 s
The solution of the differential equation is
</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T) + </span>V_capac(∞)
Since the capacitor is initially uncharged V_capac(0) = 0
And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery V_capac(∞) = 9V
This means,
V_capac (t) = (-9V)e ^(-t /T) + 9V
The charge in a capacitor is defined as Q = C*V
Where C is the capacitance and V is the Voltage across
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V = -1.87V +9V
V_capac (4.3048 s) = 7.1275 V
Q (4.3048 s) = 89.9mF*(7.1275V) = 0.6407 C
3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V
Q (∞) = 89.9mF*(9V) = 0.8091 C
<span>motion of truck constitutes of 3 travels.
1. accelerating uniformly with acceleration a1 = 2 m/s^2 until its velocity reached 20 m/s travelling a
distance of 's1' meters.
2. uniform motion with 20 m/s for a time duration t1 = 20s travelling a distance of 's2' meters.
3. uniform deceleration for t2 = 5 sec which stops the truck after travelling a distance of 's3' meters.. </span>
First calculate the time by using following equation:<span>
d= (vi+vf)/2 x t </span>
where t = time
vi is the initial velocity = 23 m/s
<span>vf is the final velocity = 0 m/s</span>
and d is the distance covered = 85m
85 = (23 + 0)/2 x t
t = 85/11.5 = 7.39s
Now use this equation to find acceleration:
a= (vf-vi)/t <span>
a = 0 – 23/ 7.39</span>
a = -3.112 m/s²
<span>so, the acceleration is 3.112 m/s</span>²<span>
</span>
Answer:
The radius r of the metal sphere.
Explanation:
From Gauss's law we know that for a spherical charge distribution with charge 
, the electrical field at distance 
from the center of the sphere is given by
What is important to notice here is that the radius of the sphere does not matter because any test charge sitting at distance feels the force as if all the charge were sitting at the center of the sphere.
This situation is analogous to the gravitational field. When calculating gravitational force due to a body like the sun or the earth, we take not of only the mass of the sun and the distance from it's center; the sun's radius does not matter because we assume all of its mass to be concentrated at the center.
To know if their research is correct or matches their hypothesis.
