Answer:
2.5 N
because Average speed is equal to distance divided by time
Answer: 2.49×10^-3 N/m
Explanation: The force per unit length that two wires exerts on each other is defined by the formula below
F/L = (u×i1×i2) / (2πr)
Where F/L = force per meter
u = permeability of free space = 1.256×10^-6 mkg/s^2A^2
i1 = current on first wire = 57A
i2 = current on second wire = 57 A
r = distance between both wires = 26cm = 0.26m
By substituting the parameters, we have that
Force per meter = (1.256×10^-6×57×57)/ 2×3.142 ×0.26
= 4080.744×10^-6/ 1.634
= 4.080×10^-3 / 1.634
= 2.49×10^-3 N/m
D chest tightness
I experienced these the past few days...
but hoped I helped ^_^
E=Fe/q
5.8x10^5N/C=Fe/(1.5x10^-9C)
Fe=(5.8x10^5N/C)(1.5x10^-9C)
Fe=8.7x10^-4N
Fe=kq1q2/r²
8.7X10^-4N=(8.99x10^9N·m²/C²)(1.5x10^-9C)(1.5x10-9C)/r²
r²=(8.99x10^9N·m²/C²)(1.5x10^-9C)(1.5x10-9C)/(8.7x10^-4N)
√r²=√0.00002325
The final answer is r=4.8x10-3m
