<span>Using Coulomb's law: k*(-0.3)*(-0.3)/(d^2)=19.2
D is the distance between the two negative charges</span>
For any temperature scale the ratio of difference of temperature will be same
so we can write
now given that in both scales the temperature must be same
<em>so at above temperature both scales will have same temperature</em>
Answer:
The height of the tower will be 35.714 m
Explanation:
We have given gauge pressure 
Density of water 
We have to find the height of the tower h
We know that gauge pressure is given by 
So the height of the tower will be 35.714 m
Answer:
option (c)
Explanation:
mass of iron = 0.10 kg
mass of copper = 0.16 kg
rise in temperature, ΔT = 35°C
specific heat of iron = 450 J/kg°C
specific heat of copper = 390 J/kg°C
Heat by iron (H1) = mass of iron x specific heat of iron x ΔT
H1 = 0.10 x 450 x 35 = 1575 J
Heat by copper (H2) = mass of copper x specific heat of copper x ΔT
H1 = 0.16 x 390 x 35 = 2184 J
Total heat H = H1 + H2
H = 1575 + 2184 = 3759 J
by rounding off
H = 4000 J
Answer:
Qsinθ/4πε₀R²θ
Explanation:
Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.
Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.
So dq = λRdθ.
Substituting dq into dE', we have
dE' = dqcosθ/4πε₀R²
= λRdθcosθ/4πε₀R²
= λdθcosθ/4πε₀R
E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ
E' = (λ/4πε₀R)[sinθ] from -θ to θ
E' = (λ/4πε₀R)[sinθ]
= (λ/4πε₀R)[sinθ - sin(-θ)]
= (λ/4πε₀R)[sinθ + sinθ]
= 2(λ/4πε₀R)sinθ
= (λ/2πε₀R)sinθ
Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ
Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ
Q = 2λRθ
λ = Q/2Rθ
Substituting λ into E', we have
E' = (Q/2Rθ/2πε₀R)sinθ
E' = (Q/θ4πε₀R²)sinθ
E' = Qsinθ/4πε₀R²θ where θ is in radians
