Answer:
The velocity of the other fragment immediately following the explosion is v .
Explanation:
Given :
Mass of original shell , m .
Velocity of shell , + v .
Now , the particle explodes into two half parts , i.e 
.
Since , no eternal force is applied in the particle .
Therefore , its momentum will be conserved .
So , Final momentum = Initial momentum
The velocity of the other fragment immediately following the explosion is v .
Explanation:
The triple beam balance is used to measure masses very precisely; the reading error is 0.05g
<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
Induced electromotive force
