Complete question:
Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.
Answer:
The electric field inside this metal resistor is 3125 V/m
Explanation:
Given;
length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m
diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m
the potential difference due to electric field between the two ends of the resistor, V = 10 V
The electric field inside this metal resistor is given by;
ΔV = EL
where;
ΔV is change in electric potential
E = ΔV / L
E = 10 / (3.2 x 10⁻³ )
E = 3125 V/m
Therefore, the electric field inside this metal resistor is 3125 V/m
It depends how you want it to work do you want to take a picture oh yea and to do that you must create a account btw
Answer:
Option B, Fix the piston in place so the volume of the pas remains constant
Explanation:
As we know

The effect on variable due to another variable can be studied by keeping the third variable constant.
Hence, in order the study the variation of temperature with pressure or vice versa, the volume needs to fixed at a certain value.
Hence, option B is correct
Answer:
1. Largest force: C; smallest force: B; 2. ratio = 9:1
Explanation:
The formula for the force exerted between two charges is

where K is the Coulomb constant.
q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.
For simplicity, let's combine Kq₁q₂ into a single constant, k.
Then, we can write

1. Net force on each particle
Let's
- Call the distance between adjacent charges d.
- Remember that like charges repel and unlike charges attract.
Define forces exerted to the right as positive and those to the left as negative.
(a) Force on A

(b) Force on B

(C) Force on C

(d) Force on D

(e) Relative net forces
In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

2. Ratio of largest force to smallest

Divide the distance traveled by the time it took:
(100 m) / (4.2 s) ≈ 23.8 m/s