Hen a gfci receptacle device is installed on a 20-ampere branch circuit (12 awg copper), what is the minimum volume allowance (i
n cubic inches) required for conductor fill for each conductor in the outlet box?
1 answer:
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Velocity:
Velocity is change in displacement with respect to time:
Analysing the units, meters (displacement) and seconds (time) are basic units:
Therefore the unit of velocity is m/s
Force:
Newton's second law of motion:
Kilogram (mass) is a basic unit, and accelerations unit can be found using the equation:
Analysing the units:
Therefore, the unit of force is:
Pressure:
Pressure is given by the equation:
where S is area of effect, F is force
Area for a basic rectangle (geometric shape is arbitrary for dimensional analysis) is found by multiplying two lengths:
, the unit of area
Dividing the aforementioned unit of force by the unit of area:
, the unit of pressure
Work:
Work is given by the equation:
, (dot product may be assumed as normal multiplication for the purposes of unit analysis)
Knowing displacement's (x) unit is m:
, the unit of work.
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct