A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.
Explanation:
From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).
To find the resistance of 260 ft (79.25 m) of size 4 AWG,
R= K * L/ A
K = 0.0214 ohm mm²/m
L = 79.25 m
A = 21.2 mm²
R = 0.0214 * 
= 0.0214 * 3.738
= 0.0792 ohm.
Thus the resistance of uncoated copper wire is 0.0792 ohm
Answer:
EH buddy use a sparkplug use a drill through a hose im from da bronx
Explanation:
Answer:
The design process is at the verify phase of Design for Six Sigma
Explanation:
In designing for Six Sigma, DFSS, is a product or process design methodology of which the goal is the detailed identification of the customer business needs by using measurements tools such as statistical data, and incorporating the identified need into the created product which in this case is the hydraulic robot Kristin Designed
Implementation of DFSS follows a number of stages that are based on the DMAIC (Define - Measure - Analyze - Improve) projects such as the DMADV which stand for define - measure - analyze - verify
Therefore, since Kristin is currently ensuring that the robot is working correctly and meeting the needs of her client the design process is at the verify phase.
Answer:
(a) 11.437 psia
(b) 13.963 psia
Explanation:
The pressure exerted by a fluid can be estimated by multiplying the density of the fluid, acceleration due to gravity and the depth of the fluid. To determine the fluid density, we have:
fluid density = specific gravity * density of water = 1.25 * 62.4 lbm/ft^3 = 78 lbm/ft^3
height = 28 in * (1 ft/12 in) = 2.33 ft
acceleration due to gravity = 32.174 ft/s^2
The change in pressure = fluid density*acceleration due to gravity*height = 78*32.174*(28/12) = 5855.668 lbm*ft/(s^2 * ft^2) = 5855.668 lbf/ft^2
The we convert from lbf/ft^2 to psi:
(5855.668/32.174)*0.00694 psi = 1.263 psi
(a) pressure = atmospheric pressure - change in pressure = 12.7 - 1.263 = 11.437 psia
(b) pressure = atmospheric pressure + change in pressure = 12.7 + 1.263 = 13.963 psia
Answer:
The distance measure from the wall = 36ft
Explanation:
Given Data:
w = 10
g =32.2ft/s²
x = 2
Using the principle of work and energy,
T₁ +∑U₁-₂ = T₂
0 + 1/2kx² -wh = 1/2 w/g V²
Substituting, we have
0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²
170 = 0.15528V²
V² = 170/0.15528
V² = 1094.796
V = √1094.796
V = 33.09 ft/s
But tan ∅ = 3/4
∅ = tan⁻¹3/4
= 36.87°
From uniform acceleration,
S = S₀ + ut + 1/2gt²
It can be written as
S = S₀ + Vsin∅*t + 1/2gt²
Substituting, we have
0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)
19.85t - 16.1t² + 3 = 0
16.1t² - 19.85t - 3 = 0
Solving it quadratically, we obtain t = 1.36s
The distance measure from the wall is given by the formula
d = VCos∅*t
Substituting, we have
d = 33.09 * cos 36. 87 * 1.36
d = 36ft
