Answer:
Explanation:
Run the code given in text file following instructions.
Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂
Answer:
The given grammar is :
S = T V ;
V = C X
X = , V | ε
T = float | double
C = z | w
1.
Nullable variables are the variables which generate ε ( epsilon ) after one or more steps.
From the given grammar,
Nullable variable is X as it generates ε ( epsilon ) in the production rule : X -> ε.
No other variables generate variable X or ε.
So, only variable X is nullable.
2.
First of nullable variable X is First (X ) = , and ε (epsilon).
L.H.S.
The first of other varibles are :
First (S) = {float, double }
First (T) = {float, double }
First (V) = {z, w}
First (C) = {z, w}
R.H.S.
First (T V ; ) = {float, double }
First ( C X ) = {z, w}
First (, V) = ,
First ( ε ) = ε
First (float) = float
First (double) = double
First (z) = z
First (w) = w
3.
Follow of nullable variable X is Follow (V).
Follow (S) = $
Follow (T) = {z, w}
Follow (V) = ;
Follow (X) = Follow (V) = ;
Follow (C) = , and ;
Explanation:
Answer:
The pressure drop is 269.7N/m^2
Explanation:
∆P = ∆h × rho × g
∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2
∆P = 0.032×860×9.8 = 269.7N/m^2