Answer:
Answer for the question is given in the attachment.
Explanation:
Answer:
11.962337 × 10^-4 N
Explanation:
Given the following :
Length L = 11.8
Charge = 29nC = 29 × 10^-9 C
Linear charge density λ = 1.4 × 10^-7 C/m
Radius (r) = 2cm = 2/100 = 0.02 m
Using the relation:
E = 2kλ/r ; F =qE
F = 2kλq/L × ∫dr/r
F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))
2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4
In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214
Hence,
(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N
Oceans and Lakes are part of the Hydrosphere.
Answer:
electrons
Explanation:
Given that,
Total charge = 9 mC = 0.009 C
0.009 C of charge passes through a wire in 3.6 s.
Let q' is the charge that passes through it in 10 s.
So,
We know that,
q = ne
Where
n is the number of electrons
So,
So, 
electrons must pass through the cross-sectional area.
Answer:
0.48
Explanation:
Given that
Frequency b1=0.4
Frequency b2=0.6
The predicted frequency of heterozygotes is given as
f(b1b2)= 2 b1 x b2
Now by putting the values
f(b1b2)= 2 b1 x b2
f(b1b2)= 2 x 0.4 x 0.6
f(b1b2)=0.48
So the answer is 0.48.
