Answer:
The unit is the barn, which is equal to 10-28 m^2 or 10-24 cm^2
Explanation:
The standard unit for measuring a nuclear cross section (denoted as σ)
Answer:
required distance is 233.35 m
Explanation:
Given the data in the question;
Sound intensity 
= 1.62 × 10⁻⁶ W/m²
distance r = 165 m
at what distance from the explosion is the sound intensity half this value?
we know that;
Sound intensity is proportional to 1/(distance)²
i.e
∝ 1/r²
Now, let r² be the distance where sound intensity is half, i.e ₂ = ₁/2
Hence,
₂/₁ = r₁²/r₂²
1/2 = (165)²/ r₂²
r₂² = 2 × (165)²
r₂² = 2 × 27225
r₂² = 54450
r₂ = √54450
r₂ = 233.35 m
Therefore, required distance is 233.35 m
The elastic potential energy stored in the stretched spring is 1 J.
<h3>What is Hooke's law?</h3>
Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.
Given that;
Force on the spring = 350 Newton
Distance stretched = 7 centimeters or 0.07 m
Hence;
F = ke
k = F/e = 350 Newton/0.07 m = 5000 N/m
Work done in stretching a spring = 1/2ke^2
= 0.5 × 5000 × (2 × 10^-2)^2 =1 J
Learn more about elastic potential energy: brainly.com/question/156316
Answer:
Explanation:
We have to find electric potential V at a distance r.
a) For r>R,
The electric field in the cylinder is given by
E.A equating it to the other electric field given by
б.A/ε₀
Here the area of cylinder is given by= 2*3.14*r*L
While for the outside, the area= 2*3.14*R*L
Equating both, we get
E= бR/rε₀
Now,
The potential difference is given as:
ΔV= -бR/rε₀ and integrating right side with respect to dr under limits r and R.
Where ΔV= V₀-V
So solving we get
V₀=V-бR/ε₀ln (r/R)
b) For r<R i.e. inside the cylinder
There will be no electric field produced as E=0
So ultimately Vin= V
c) V=0 at r= infinity.
