<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
Among the choices the one that most likely cause a sound wave is C or the a spoon hitting the side of a bowl while mixing <span>because that's going to make a sound.
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
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The east bound train travels at a speed of 95 mile per hour and the west bound train at 75 miles per hour.
Assuming eastbound train will cover distance x then the west bound train will cover distance 272-x .
Therefore, since time taken will be the same then; x/95 = (272-x)/75
= 75x = 95 (272-x)
= 75x = 25840 - 95x
= 170 x= 25840
x = 152 miles
Thus time taken will either be x/95 or (272-x)/75
= 152/95
= 1.6 hours or 1 hour 36 minutes
Answer: The planet, named after the Roman goddess of art and beauty, can actually be bright enough to cast shadows on a moonless night. It appears so close to the sun because its orbital radius is smaller than the Earth's, and because it also moves faster than Earth, its orbital period is shorter.
