Answer:
This is because normal force is exerted perpendicularly to the point of contact between the upper and lower objects.
Explanation:
This is because the upper object is still subject to gravitational pull. Therefore, the amount of force it exerts on the lower object due to gravity will be equal to the normal force that acts in the negative direction of gravitational force. Additionally, normal force is evident because the upper object will not go into the lower object.
<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
Answer:
$6.68
Explanation:
The current running through the device would be its outlet voltage divided by the resistance:
I = U/R = 115 / 11.4 = 10.09 A
The power generated by this device is the product of its current and voltage
P = IU = 10.09 * 115 = 1160 W or 1.16 kW
If running this device for 36 hours then the total energy it would consume is
E = Pt = 1.16 * 36 = 41.76 kWh
Therefore the total cost of electrical energy is
C = Ec = 41.76*0.16 = $6.68
260 meters im pretty sure because rate*time=distance