If there are 3 resistors of .5 ohms, 1 ohms, and 1 ohms connected in parallel then what is the equivalent resistance
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The number of oscillations of the toy in a second is 1.25.
<h3>What is frequency?</h3>- This is the number of complete oscillation of an object is a given period.
The given parameter:
- Frequency of the toy, F = 1.25 Hz
The frequency of an object is calculated as follows;
where;
- n is the number of oscillations
- t is the time of motion
The number of oscillations of the toy in a second is calculated as follows;
Learn more about frequency here: brainly.com/question/10728818
Answer:
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Also it is given that
q1=-0.7uC
q4=-1.7uC
q3=-1.7uC
Also the distance are given as
a=2.2cm=0.022m
d2=3.6cm=0.036m
Explanation:
The potential energy due to point R is given as
The potential energy due to charge q1 and q3 plus the potential energy due to charge q4 and q1 plus the potential energy due to charge q3 and q4
So, let take it one after the other
Potential energy is give as
P.E=kq1q2/r
Therefore,
Potential energy due to charge q1 and q3
U¹³=kq1q3/r
To get the distance between charge q1 and q3, we will apply Pythagoras theorem
r=√(d2²+a²)
r=√(0.036²+0.022²)
r=0.0422m
k is a constant =9×10^9Nm²/C²
Then,
U¹³=kq1q3/r
U¹³=9×10^9×0.7×10^-6×1.7×10^-6/0.0422
U¹³=0.254J
Potential energy due to charge q1 and q4
U¹⁴=kq1q4/r
To get the distance between charge q1 and q4, we will apply Pythagoras theorem
r=√(d2²+a²)
r=√(0.036²+0.022²)
r=0.0422m
k is a constant =9×10^9Nm²/C²
Then,
U¹⁴=kq1q4/r
U¹⁴=9×10^9×0.7×10^-6×1.7×10^-6/0.0422
U¹⁴=0.254J
Potential energy due to charge q3 and q4
U³⁴=kq3q4/r
r=2a=2×0.022=0.044m
k is a constant =9×10^9Nm²/C²
Then,
U³⁴=kq3q4/r
U³⁴=9×10^9×1.7×10^-6×1.7×10^-6/0.044
U¹⁴=0.591J
Then, the total energy is
U= U¹³+ U¹⁴ + U³⁴
U=0.254+0.254+0.591
U=1.099J
Then also, the potential energy is zero because at infinity both U¹³ and U¹⁴ will have infinite potential because their distance apart will be infinite.
