Answer:
E = k Q 1 / (x₀-x₂) (x₀-x₁)
Explanation:
The electric field is given by
dE = k dq / r²
In this case as we have a continuous load distribution we can use the concept of linear density
λ= Q / x = dq / dx
dq = λ dx
We substitute in the equation
∫ dE = k ∫ λ dx / x²
We integrate
E = k λ (-1 / x)
We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁
E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)
E = k λ (x₂ -x₁) / (x₀-x₂) (x₀-x₁)
We replace the density
E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]
E = k Q 1 / (x₀-x₂) (x₀-x₁)
Answer:
0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.
Explanation:
First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:
P₁ = F/A
where,
P₁ = Pressure exerted by empty rack = ?
F = Force exerted by empty rack = Weight of Empty Rack = 40 lb
A = Base Area = 452.4 in²
Therefore,
P₁ = 40 lb/452.4 in²
P₁ = 0.088 psi
Now, we calculate the pressure exerted by the rack along with the coat.
P₂ = F/A
where,
P₂ = Pressure exerted by rack filled with coats= ?
F = Force exerted by filled rack = Weight of Filled Rack = 65 lb
A = Base Area = 452.4 in²
Therefore,
P₂ = 65 lb/452.4 in²
P₂ = 0.144 psi
Now, the difference between both pressures is:
ΔP = P₂ - P₁
ΔP = 0.144 psi - 0.088 psi
<u>ΔP = 0.056 psi</u>
<h3><u>Answer</u>;</h3>
= 22°
<h3><u>Explanation</u>;</h3>
- According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The constant value is called the refractive index of the second medium with respect to the first.
- Therefore; Sin i/Sin r = η
In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33
Thus;
Sin 30 / Sin r = 1.33
Sin r = Sin 30°/1.33
= 0.3759
r = Sin^-1 0.3759
= 22.08
<u>≈ 22°</u>
Brian’s Complexity Brian’s Complexity Brian’s Complexity Brian’s Complexity
Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B
Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.
Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.
Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)
Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb
Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb
Answer:

where