Answer: A solution is the end of a task.
Explanation:
Explanation:
Gauss Law relates the distribution of electric charge to the resulting electric field.
Applying Gauss's Law,
EA = Q / ε₀
Where:
E is the magnitude of the electric field,
A is the cross-sectional area of the conducting sphere,
Q is the positive charge
ε₀ is the permittivity
We be considering cases for the specified regions.
<u>Case 1</u>: When r < R
The electric field is zero, since the enclosed charge is equal to zero
E(r) = 0
<u>Case 2</u>: When R < r < 2R
The enclosed charge equals to Q, then the electric field equals;
E(4πr²) = Q / ε₀
E = Q / 4πε₀r²
E = KQ /r²
Constant K = 1 / 4πε₀ = 9.0 × 10⁹ Nm²/C²
<u>Case 3</u>: When r > 2R
The enclosed charge equals to Q, then the electric field equals;
E(4πr²) = 2Q / ε₀
E = 2Q / 4πε₀r²
E = 2KQ /r²
Answer:
Explanation:
At the bottom the tension would be upwards and the weight downwards, their difference being the centripetal force. Taking the upwards direction as positive we then have:
where we have used the equation for centripetal acceleration. Thus we have:
Answer:
Explanation:
The diffraction pattern is given as
Sinθ = mλ/ω
Where m=1,2,3,4....
Now, when m=1
Sinθ = λ/ω
Then,
ω = λ/Sinθ
The width of the central bright fringe is given as
y=2Ltanθ. From trigonometric
Then,
θ=arctan(y/2L)
Given that,
y=0.052m
L=0.55m
θ=arctan(0.052/2×0.55)
θ=arctan(0.0473)
θ=2.71°
Substituting this into
ω = λ/Sinθ
Since λ=544nm=544×10^-9m
Then,
ω = 544×10^-9/Si.2.71
ω = 1.15×10^-5m.
Answer:
There are 8345714 cell membranes.
Explanation:
It is given that,
The thickness of a cell membrane is 7 nm
We need to find the number of cell membranes would it take to make a stack 2.3 inches high.
Firstly, the units must be same i.e. converting 2.3 inches to m
1 inch = 0.0254 m
2.3 inches = 0.05842 m
Let there are n number of cell membranes. So,
or
n = 8345714 cell membranes
