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4 years ago

A Vector is a quantity that has magnitude, direction, and sense. a)-True b)- False

Engineering

1 answer:

Ulleksa [173]4 years ago

3 0

The answer would be - b false because it doesn’t have sense

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Technician A says that a seal can be pried out of a bore using a sharp chisel. Technician B says that smaller metal-backed seals

attashe74 [19]

Answer:

The correct answer is letter "C": Both.

Explanation:

Industrial seals are used at interfaces between components to prevent leakage, to maintain heat, and to avoid contamination. The design, construction, and materials they use vary depending on industrial use but the most common are Polytetrafluoroethylene (PTFE), Nitrile Buna Rubber (NBR), and fluorocarbon.

Thus, using a sharp chisel could pry a seal out of a hole and a regular socket can often be used to force smaller metal-backed seals into place. Thus, technicians "A" and "B" are correct.

6 0

3 years ago

Read 2 more answers

Write the closed-loop transfer function for the system and determine the closed-loop pole locations. Is the closed-loop system s

Naily [24]

Answer:

Check the explanation

Explanation:

To write the closed-loop transfer function for the system and determine the closed-loop pole locations as well as to determine the damping ratio and the natural frequency, we will be be doing a step by step calculation and explanation including the graphical presentation as seen in the attached images below.

6 0

4 years ago

Consider 4.8 pounds per minute of water vapor at 100 lbf/in2, 500 oF, and a velocity of 100 ft/s entering a nozzle operating at

andriy [413]

Answer:

A) $v_2 = 2016.80 ft/s$

B) $\Delta s = 0.006 Btu/lbm R$

Explanation:

Given data:

P-1 = 100 lbf/in^2

$T_1 = 500$ degree f

$V_1 = 100 ft/s$

$P_2 = 40 lbf/inc^2$

effeciency = 80%

from steady flow enerfy equation

$h_1 +\frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}$

where h1 and h2 are inlet and exit enthalpy

for P1 = 100 lbf/in^2 and T1 = 500 degree F

$H_1 = 1278.8 Btu/lbm$

$s_1 = 1.708 Btu/lbm -R$

for P1 = 40 lbf/in^2

$H_1 = 1193.5 Btu/lbm$

$s_1 = 1.708 Btu/lbm -R$

exit enthalapy h_2

$\eta = \frac{h_1 - h'_2}{ h_1 - h_2}$

$0.80 = \frac{1278.8 - h'_2}{1278.8 -1193.5} = 1197.77 Btu/lbm$

from above equation

$1278.8 \times 25037 + \frac{100^2}{2} = 1197.77 \times 25037 + \frac{v_2^2}{2}$ [1 Btu/lbm = 25037 ft^2/s^2]

$v_2 = 2016.80 ft/s$

b) amount of entropy

$\Delta s = s_2 - s_1$

$s_1 = 1.708 Btu/lbm -R$

at $h_2 = 1197.77 Btu/lbm [\tex] and [tex]P_2 = 40 lbf/in^2$

$s_2 is 1.714 Btu/lbm -R$

$\Delta s = 1.714 - 1.708 = 0.006 Btu/lbm R$

6 0

3 years ago

While walking across campus one windy day, an engineering student speculates about using an umbrella as a "sail" to propel a bic

makvit [3.9K]

Answer:

$Given data:\\While walking across campus one windy day\\Frontal area, $A=0.3 m ^{2}$\\Wind speed $V=24 Km / hr$\\The drag coefficient $C_{D, b}=1.2$\\The combined mass $m=75 kg$\\Umbrella diameter, $D=1.22 m$\\Velocity of wind $V=24 \frac{ km }{ hr }$\\The rolling resistance $C_{R}=0.75 \%$$

Solution:

Note: Refer the diagram

$Basic equation:\\'s law of motion: $\sum F_{x}=m a_{x}$\\Lift coefficient, $C_{L}=\frac{F_{L}}{\frac{1}{2} \rho V^{2} A_{p}}$\\Drag coefficient, $C_{D}=\frac{F_{D}}{\frac{1}{2} \rho V^{2} A_{p}}$$

$From force balance equation:\\$\sum F_{x}=F_{D}-F_{R}=0$\\But $F_{D}=\left(C_{D, \alpha} A_{u}+C_{D, B} A_{b}\right) \frac{1}{2} \rho\left(V_{\nu}-V_{b}\right)^{2}$$F_{R}=C_{R} m g$\\Area of the Umbrella $A_{u}=\frac{\pi D_{u}^{2}}{4}$$A_{x}=\frac{\pi \times 1.22^{2}}{4} m ^{2}$$A_{v}=1.17 m ^{2}$$

Drag coefficient data for selected objects table at

Hemisphere (open end facing flow), $C_{D, x}=1.42$

Substituting all parameters,

$\begin{aligned}&F_{R}=0.0075 \times 75 \times 9.81\\&F_{R}=5.52 N\end{aligned}$

Then,

$\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}$

And the equation becomes,

$\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}$