For balancing the lever, force on both the sides shall be equal. so,
Force on 3 m end = m × a = 3 × 98.1 = 294.3
Now, on 6 m end, it would be: = 294.3/6 = 49.05
After rounding-off to the nearest hundredth value, it would be: 49 N
Finally, Option A would be your correct answer.
Hope this helps!
Answer: C) The two particles will move away from each other
Explanation:
When two electrically charged bodies come closer, appears a force that attracts or repels them, depending on the sign of the charges of this two bodies.
This is stated by Coulomb's Law:
"The electrostatic force 
between two point charges 
and 
is proportional to the product of the charges and inversely proportional to the square of the distance 
that separates them, and has the direction of the line that joins them"
Mathematically this law is written as:
Where 
is a proportionality constant.
Now, if and have the same sign charge (both positive or both negative), a repulsive force will act on these charges.
Answer:
Explained
Explanation:
Michelson contrast is used for patterns where the distribution of bright and dark segments is nearly equal.
It is given by:
where I_max = maximum illumination and I_min = minimum illumination
we know that
typically, I_min = 54% of I_max (general standard)
or I_min = 0.54 I_max
putting this value in above equation to get m
this approximately corresponds to m = 0.3 or 30%
hence, 30% recommended as the minimum Michelson contrast
<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
