a=Δv/Δt=(v-v₀)/t=(46.1-18.5)/2.47=27.6/2.47=11.17 m/s²
d=v₀t+at²/2=18.5*2.47+11.17*2.47²/2=45.69+34.08=79.776 m
<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
R = ρl/A
From the equation above R = Resistance, l = length, A = Cross Sectional Area of wire.
From the equation, it can be seen that R would increase if the wire's area is reduced.
If the area of the wire is reduced, means the same thing as:
<span>A. decreasing the wire’s thickness</span>
Answer: 4.44302176697 m/s^2
Explanation:
formula for acceleration is (final velocity - initial velocity)/time. So in your case:
(34.7m/s - 0m/s)/ 7.81s = 4.44302176697 m/s^2
Answer:
lol do you still need help?
Explanation: