215
I am assuming that those are the width and length. So to find area all you have to do is multiply the two because lw = a. 12 1/2*17 1/5 = 215
25/2*86/5 = 2150/10 = 215
Answer:
Coil 2 have 235 loops
Explanation:
Given
The number of loops in coil 1 is n
₁=
159
The emf induced in coil 1 is ε
₁
=
2.78
V
The emf induced in coil 2 is ε
₂
=
4.11
V
Let
n
₂ is the number of loops in coil 2.
Given, the emf in a single loop in two coils are same. That is,
ϕ
₁/n
₁=
ϕ
₂
n
₂⟹
2.78/159
=
4.11/
n
₂
n₂=
n₂=235
Therefore, the coil 2 has n
₂=
235 loops.
Thw question is not complete. The complete question is;
Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC
Answer:
Option C: E = 75 N/C
Explanation:
We are given;
Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m
Distance on the y-axis; d = 1.6 m
Now, the formula for electric field with uniform linear density is given as;
E = λ/(2•π•r•ε_o)
Where;
E is electric field
λ is uniform linear density = 6.7 × 10^(-9) C/m
r is distance = 1.6m
ε_o is a constant = 8.85 × 10^(-12) C²/N.m²
Thus;
E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))
E = 75.31 N/C ≈ 75 N/C
A. Plasma
That is your answer!
Answer:
The tensile stress on the wire is 550 MPa.
Explanation:
Given;
Radius of copper wire, R = 3.5 mm
extension of the copper wire, e = 5.0×10⁻³ L
L is the original length of the copper wire,
Young's modulus for copper, Y = 11×10¹⁰Pa.
Young's modulus, Y is given as the ratio of tensile stress to tensile strain, measured in the same unit as Young's modulus.
Therefore, the tensile stress on the wire is 550 MPa.
