Answer:Resistor-Capacitor (RC) circuits, when driven by a voltage/current source, display a type of time-dependent charging or discharging since the charge from the capacitor goes through the resistor. When considering a discharging phase, the time-dependent voltage takes the form
Explanation:
Vt = Voe^-t/RC
Vt equal time dependent voltage
R is the resistance
C is the capacitance of the oscilloscope
Given
Area of capacitor equal to 0.1 * 0.002m
Distance between plates equal 1/1000 = 0.0001m
Vo = 100v supply voltage
Resistance R =:1000ohms
Vt time dependent voltage = 50V
To find capacitance C = Eo(A/d)
C = (8.85 * 10^-12)* (0.002/0.001)
C= 1.77 * 10^-11
Solving the equation Vt = Voe^-t/RC
t = 0.6931RC
t = 0.6931(1000)(1.77*10^-11)
t = 1.2*10^-9
t = 1.2ns
It will take the oscilloscope 1.2ns to reach 50V
Answer:
C) 4.0026 amu
Explanation:
Helium (He) is the 2nd element on the Periodic Table.
It's neutral atom has 2 protons and 2 electrons.
It is in the 1st period and the 18th row.
It is also a Noble gas.
It will be unstable system because it will not be able to recover from the disturbance
Answer:Principle of rectilinear propagation of light
Explanation:Principle of rectilinear propagation of light
Rectilinear propagation of light refers to the propensity of light to travel along a straight line without any interference in its trajectory. ... It is because light travels along a straight line and leaves only the areas where the object interferes.
Answer:

Explanation:
The force on the point charge q exerted by the rod can be found by Coulomb's Law.

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.
In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.
We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.
Applying Coulomb's Law:

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.
Now, we have to write 'dq' in term of the known quantities.

Now, substitute this into 'dF':

Now we can integrate dF over the rod.
