<h2>
Answer:</h2>
1.68 x 10⁻⁸Ωm
<h2>
Explanation:</h2>
The resistance (R) of a wire is related to its length(L), its material resistivity(ρ) and its crossectional area(A) as follows;
R = ρL/A ------------------------(i)
Where;
A = πd² / 4 [where d = diameter of the wire]
From the question;
L = 6.90m
d = 2.15mm = 0.00215m
R = 0.0320Ω
First calculate the crossectional area (A) of the wire as follows;
A = πd² / 4
[Take π = 3.142]
d = 0.00215m
∴ A = 3.142 x (0.00215)² / 4
∴ A = 0.000003631m²
Now, substitute the values of A, L, and R into equation (i) as follows;
R = ρL/A
0.0320 = ρ x 6.90 / 0.000003631
0.0320 = 1900302.95 x ρ
Solve for ρ;
=> ρ = 0.0320 / 1900302.95
=> ρ = 1.68 x 10⁻⁸Ωm
Therefore, the resistivity of the material of the wire is 1.68 x 10⁻⁸Ωm
69 i agree with her hope this helps
Answer:
d) 2Fr
Explanation:
We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².
We now evaluate the integral from r = +r to -r
W = ∫Fdr
= ∫(-e²/4πε₀r²)dr
= -∫e²dr/4πε₀r²
= -e²/4πε₀∫dr/r²
= -e²/4πε₀ × -[1/r] from r = +r to -r
W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.
Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.
So W = -2e²/4πε₀r = 2Fr.
So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr
The ball rises for v/g seconds; which equals 14.7/9.8=1.5 seconds . After this time, it’s height will be:
h(t)=g/2(1.5)²+14.7(1.5)
=-4.9 x 2.25 + 22.05
=11.025m
The ball then falls for 49+11.025=60.025m, which takes:
g/2t²=60.025
t²=12.25
t=3.5 secs
Total time: 1.5+3.5=5 seconds
