We have that The ratio U1/U2 of their potential energies due to their interactions with Q is
From the question we are told that
Question 1
Charge q1 is distance r from a positive point charge Q.
Question 2
Charge q2=q1/3 is distance 2r from Q.
Charge q1 is distance s from the negative plate of a parallel-plate capacitor.
Charge q2=q1/3 is distance 2s from the negative plate.
Generally the equation for the potential energy is mathematically given as

Therefore
The Equations of U1 and U2 is
For U1

For U2

Since
U is a function of q and q2=q1/3
Therefore

For Question 2
For U1

Therefore

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<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:
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Answer: there are 15 coins of $2 and 18 coins of $5
Explanation:
I will answer in English.
X is the number of $5 coins.
Y is the number of $2 coins.
We have the system of equations:
Y + X = 33
Y*2 + X*5 = 120
first, we must isolate one of the variables in one of the equations and then replace it in the other equation, let's isolate Y in the first equation:
Y = 33 - X.
Then we can replace it in the other equation:
(33 - X)*2 + X*5 = 120
66 - X*2 + X*5 = 120
X*3 = 54
X = 54/3 = 18
and using the equation for Y.
Y = 33 - X = 33 - 18 = 15
So there are 15 coins of $2 and 18 coins of $5
Answer:
1.0125 x 10^19
Explanation:
current flowing through conductive wire= 9mA = 9 x 10^ -3 A
charge passing per 3 min
Q = It
= 9 x 10^ -3 x (3 x 60)
= 1.620 C
no of electrons in charge
Q = ne
1.620 = n x 1.6 x 10 ^ -19
n. = 1.0125 x 10 ^19
Magnetic flux can be calculated by the product of the magnetic field and the area that is perpendicular to the field that it penetrates. It has units of Weber or Tesla-m^2. For the first question, when there is no current in the coil, the flux would be:
ΦB = BA
A = πr^2
A = π(.1 m)^2
A = π/100 m^2
ΦB = 2.60x10^-3 T (π/100 m^2 ) ΦB = 8.17x10^-5 T-m^2 or Wb (This is only for one loop of the coil)
The inductance on the coil given the current flows in a certain direction can be calculated by the product of the total number of turns in the coil and the flux of one loop over the current passing through. We do as follows:
L = N (ΦB ) / I
L = 30 (8.17x10^-5 T-m^2) / 3.80 = 6.44x10^-4 mH