Q: Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two charges q1'=2q1 and q2'=3q2 that that are distance d apart?
Answer:
The force = 38.4 N
Explanation:
From coulombs law,
F = kq₁q₂/r² ............................ Equation 1
Where F = Force of attraction or repulsion between the charges, q₁ and q₂ = first and second charge respectively, r = distance between the charges, k = constant of proportionality.
When, F = 6.4 N, r = d m.
6.4 = kq₁q₂/d²......................... Equation 1
When q₁' = 2q₁, q₂' = 3q₂, r = d cm
F = k(2q₁)(3q₂)/d²
F = 6kq₁q₂/d².......................... Equation 2
Dividing Equation 1 by equation 2
6.4/F = kq₁q₂/d²/(6kq₁q₂/d²)
6.4/F = 1/6
F = 6.4×6
F = 38.4 N.
Thus the force = 38.4 N
Answer:
showm
Explanation:
Consider a dipole having magnetic moment 'm' is placed in magnetic field 
then the torque exerted by the field on the dipole is
Now to rotate the dipole in the field to its final position the work required to be done is
Minimum energy mB is for the case when m is anti parallel to B.
Minimum energy -mB is for the case when m is parallel to B.
1,119,000 basically 746 times 150 then multiply that answer by 10
Answer:
See attached document
Explanation:
Entire process for deriving the asked expression dV across the bridge as function of dP is illustrated in the attachment below.
The document gives a step-by step process for arriving at the expression. However, manipulation of algebraic equations is skipped for the conciseness of the document.
It also gives the expression for the case when all resistors have different nominal values.
Answer:
I do belive that it is B hrs cn I an gn
