Answer:
Explanation:
Initially no of atoms of A = N₀(A)
Initially no of atoms of B = N₀(B)
5 X N₀(A) = N₀(B)
N = N₀ 
N is no of atoms after time t , λ is decay constant and t is time .
For A
N(A) = N(A)₀ 
For B
N(B) = N(B)₀ 
N(A) = N(B) , for t = 2 h
N(A)₀ = N(B)₀
N(A)₀ = 5 x N₀(A)
= 5
= 5 
half life = .693 / λ
For A
.77 = .693 / λ₁
λ₁ = .9 h⁻¹
= 5
Putting t = 2 h , λ₁ = .9 h⁻¹
= 5 
= 30.25
2 x λ₂ = 3.41
λ₂ = 1.7047
Half life of B = .693 / 1.7047
= .4065 hours .
= .41 hours .
It’s d energy because it’s referring to power changing because the power unit is (J/s) joule per second known as watt which is the same as implieing power
Answer:
same value in R and 2R E = E₀ = σ / 2ε₀
Explanation:
For this exercise we use Gauss's law
Ф = E. dA = 
/ε₀
We define a Gaussian surface with a cylinder with the base being parallel to the load sheet, so the electic field line and the normal line to the base are parallel and the scalar product is reduced to the algebraic product, in the parts the angle is 90º and the dot product is zero
As the sheet has two faces
2E A = q_{int} /ε₀
The charge inside the cylinder is
σ = q_{int} / A
q_{int} = σ A
We substitute
E = σ / 2ε₀
We see that this expression is independent of the distance, so it has the same value in R and 2R
E = E₀ = σ / 2ε₀
Answer:
8 x 10⁻⁷ x I / r
Explanation:
Two parallel long wires are carrying current I . Let the direction be towards the right in the farthest and towards the left in the nearest. Magnetic field due to current I at a distance d is given by the expression
B = μ₀ 2 I / 4π d
I the present case distance d = r/2
Magnetic field due to one wire at point d = r/2 is
B₁ = μ₀ 2 I / (4π r / 2 )
= 10⁻⁷ x 4I / r
Magnetic field due to the other wire at point d = r/2 is
B₂ = μ₀ 2 I / (4π r / 2 )
= 10⁻⁷ x 4I / r
Direction of magnetic field due to both the wires at the mid point P will be same . It will be in downward direction in the given scenario
So total magnetic field
B = B₁ + B₂
= 2 x 10⁻⁷ x 4I / r
= 8 x 10⁻⁷ x I / r
