Answer:
F1 = K Q1 Q2 / R1^2
F2 = K Q1 / 2 * Q2 / (2 R1)^2
F2 / F1 = 1/2 / 4 = 1/8
The new force is 5N (1/2 due to charge and 1/4 due to distance)
A square loop whose sides are long is made of copper wire of radius , given the resistivity of copper is . if the magnetic field perpendicular to the loop changes at a constant rate of I = 14.029 mA.
The basic characteristic of a substance that measures how effectively it resists an electric current is called electrical resistance. A material with low resistance is a material that easily conducts electric current. A Greek letter is often used to indicate resistivity. Electrical resistance is a basic property of a material that measures how strongly it resists an electric current. The SI unit for electrical resistance is the ohmmeter.
We use magnetic field as a tool to describe how the magnetic field is distributed in the space around and inside something of a magnetic nature. A material with low resistance is a material that easily conducts electric current. A Greek letter is often used to indicate resistivity. An ohmmeter is a unit of electrical resistance in the SI system.
Learn more about magnetic field here;
brainly.com/question/24397546
#SPJ4
The complete question is :
A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?
Answer:
The potential difference between the ends of a wire is 60 volts.
Explanation:
It is given that,
Resistance, R = 5 ohms
Charge, q = 720 C
Time, t = 1 min = 60 s
We know that the charge flowing per unit charge is called current in the circuit. It is given by :
I = 12 A
Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :
V = IR
V = 60 Volts
So, the potential difference between the ends of a wire is 60 volts. Hence, this is the required solution.
Answer:
t = 13.7 s or t = 14 s with proper significant figures
Explanation:
The initial speed is 0 m/s since the car starts from rest, acceleration is 5.5 m/s2 and distance is 523 m.
Since we have initial speed, acceleration and distance we can use the following formula to find the time. We can now use algebra to work out our answer.
d = vt +
at²
523 = (0)t + (
)(5.5)t²
523 = 2.8t²
186.8 = t²
13.7 s = t
(t = 14 s with proper significant figures)