The local charging of the comb is due to induction. The negative object's charge pushes electrons in the comb away from the side close to the charged object causing that part to be positively charged. Note that the comb's net charge is still zero provided it doesn't touch the object. When you move the comb away its electrons will redistribute.
Answer:
<h2> 27m/s</h2>
Explanation:
Given data
initital velocity u=15m/s
deceleration a=3m/s^2
time t= 4 seconds
final velocity v= ?
Applying the expression
v=u+at------1
substituting our data into the expression we have
v=15+3*4
v=15+12
v=27m/s
The velocity after 4 seconds is 27m/s
According to Newton's 2nd law of motion:
F = m * a where F is the force applied in Newtons, m is the mass of the object in kg, and a is the acceleration of the object in m/

.
Therefore the force applied in this situation is simply:
F = 6 kg * 2.3 m/

= 13.8 N
Hope this helps!
<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