All categories

3 years ago

Unit 6 - Quiz 1

Engineering

1 answer:

AfilCa [17]3 years ago

4 0

Answer:

c is the correct answer

You might be interested in

The state of plane strain on an element is:

balu736 [363]

Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

θ= -13.28° and 76.72°

To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain

$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

$\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )$

ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

To determine the direction of γmax

$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

= 1.67 *10^-4

4 0

4 years ago

Give me source code of Simple openGL project. ( without 3D or Animation) simple just.

Ivan

Answer:

Use GitHub or stackoverflow for this answer

Explanation:

It helps with programming a lot

4 0

3 years ago

Assume the impedance of a circuit element is Z = (3 + j4) Ω. Determine the circuit element’s conductance and susceptance.

djyliett [7]

Answer:

B. G = 333 mS, B = j250 mS

Explanation:

impedance of a circuit element is Z = (3 + j4) Ω

The general equation for impedance

Z = (R + jX) Ω

where

R = resistance in ohm

X = reactance

R = 3Ω X = 4Ω

Conductance = 1/R while Susceptance = 1/X

Conductance = 1/3 = 0.333S

= 333 mS

Susceptance = 1/4 = 0.25S

= 250mS

The right option is B. G = 333 mS, B = j250 mS

8 0

3 years ago

A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027

Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = $\frac{d}{2}$ = 3 inches

density, $\rho}$ = 493 lbm/ $ft^{3}$

S.G = 1.0027

g = 9.8 m/ $m^{2}$ = 386.22 inch/ $s^{2}$

Solution:

Using the formula for terminal velocity,

$v_{T}$ = $\sqrt{\frac{2V\rho g}{A \rho C_{d}}}$ (1)

$(Since, m = V\times \rho)$

where,

V = volume of sphere

$C_{d}$ = drag coefficient

Now,

Surface area of sphere, A = $4\pi r^{2}$

Volume of sphere, V = $\frac{4}{3} \pi r^{3}$

Using the above formulae in eqn (1):

$v_{T}$ = $\sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho g}{4\pi r^{2} \rho C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2gr}{3C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}$

Therefore, terminal velcity is given by:

$v_{T}$ = $\frac{27.79}{\sqrt{C_d}}$ inch/sec

3 0

4 years ago

In order to test the reverse route back towards the original host, which of the following will you use? A Standard ping B Extend

Vlad1618 [11]

The function that you will use in order to test the reverse route back toward the original host is known as Extended ping. Thus, the correct option for this question is B.

<h3>What is Ping and traceroute for?</h3>

Ping and traceroute are the common commands you can effectively utilize in order to troubleshoot network problems. Ping is a simple command that can typically test the reachability of a device on the network. While traceroute is a command you use to 'trace' the route that a packet takes when traveling to its destination.

Extended ping permits a router's ping command to use The router's LAN IP address from within the subnet, fully testing the route back to the subnet. A standard ping often does not test the reverse route that you need to back toward the original host.

Therefore, the function that you will use in order to test the reverse route back toward the original host is known as Extended ping. Thus, the correct option for this question is B.

To learn more about Ping and traceroute, refer to the link:

brainly.com/question/28333920

#SPJ1

6 0

1 year ago