Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer:
U = 1 / r²
Explanation:
In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related
F = - dU / dr
this derivative is a gradient, that is, a directional derivative, so we must have
dU = - F. dr
the esxresion for strength is
F = B / r³
let's replace
∫ dU = - ∫ B / r³ dr
in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product
let's evaluate the integrals
U - Uo = -B (- / 2r² + 1 / 2r₀²)
To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)
U = B / 2r²
we substitute the value of B = 2
U = 1 / r²
Answer:
Explanation:
separation between two gaps, d = 5 cm
angle between central and second order maxima, θ = 0.52°
use
d Sinθ = n λ
n = 2
0.05 x Sin 0.52° = 2 x λ
λ = 2.27 x 10^-4 m
λ = 226.9 micro metre
Answer:
power requirement is 23.52 × 
W
Explanation:
given data
flow rate q = 2 m³/s
elevation h = 1200 m
density of the water ρ = 1000 kg/m³
to find out
power requirement
solution
we will get power by the power equation that is
power = ρ× Q× g× h ...................1
put here all value we get power
power = ρ× Q× g× h
power = 1000 × 2 × 9.8 × 1200
power = 23.52 ×
so power requirement is 23.52 × W
Explanation:
Below is an attachment containing the solution.
