three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650
By applying Newton's second law of motion;
ma = mg - T
Where,
m = mass; a = downward accelerations (+ve value) or upward acceleration (-ve value); g = gravitational acceleration; T = tension.
For the current case, the velocity is constant therefore,
a = 0
Then,
0 = mg - T
T = mg = 115*9.81 = 1128.15 N
Tension in the cable is 1128.15 N.
Answer:
Explanation:
From the information given,
V = 4 volts
A = 2 amps
a)
In the first instance, one light on the sting goes out and the whole string of lights no longer turns on. This means that the circuit is a series circuit.
b) Total voltage = 4 x 100 = 400 V
The current passing through each bulb is the same. Thus
Total Current = 2 Amps
Recall, V = IR
R = V/I
Thus,
Resistance = 400/2
Resistance = 200 ohms
c) In this case, one light goes out on the string all other lights still turn on. This means that the circuit is a parallel circuit
d)
The voltage is the same
Total voltage = 4 volts
Total current = 2 x 100 = 200 amps
Total resistance = 4/200
Total resistance = 0.02 ohms
Answer:
v = 10 V and E = 2 10³ N/C
Explanation:
The electrical potentials and the electric field at one point are related by the expression
ΔV = - ∫ E. dS
Where the bold indicates vector quantities, E is the electric field and S is the line of displacement of the load, in general displacement is perpendicular to the equipotential lines, which reduces the product scales to the ordinary product.
If the potential difference is the most usual that is V = 10 V, the electric field is
s = 0.5 cm = 0.5 10⁻² m
E = ΔV / S
E = 10/0.5 10⁻²
E = 2 10³ N / C
Answer:
The frequency of the signal is 2 GHz
Explanation:
Given;
period of the clock signal, T = 500 ps = 500 x 10⁻¹² s
the frequency of the signal is given by;
F = 2 GHz
Therefore, the frequency of the signal is 2 x 10⁹ Hz or 2 GHz
