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3 years ago

Select the correct answer

Engineering

1 answer:

kifflom [539]3 years ago

5 0

Answer: C. Exchanging ideas with the group will help build his knowledge and reputation

Explanation:

Professional networking refers to the building of professional relationships. This helps in establishing connections that are mutually-beneficial with the people that are in ones profession.

Based on the scenario given, the reason for Vijay to join a networking group is that exchanging ideas with the group will help build his knowledge and reputation.

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The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.

zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft $J\tau$

shaft $J\tau$ = $\dfrac{\pi}{2}r^4$

where ;

r = 1 in /2

r = 0.5 in

shaft $J \tau$ = $\dfrac{\pi}{2} \times 0.5^4$

shaft $J\tau$ = 0.098218

Now; the angle of twist at B with respect to D is calculated by using the expression

$\phi_{B/D} = \sum \dfrac{TL}{JG}$

$\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

where;

$T_{CD} \ \ and \ \ L_{CD}$ are the torques at segments CD and length at segments CD

${T_{BC} \ \ and \ \ L_{BC}}$ are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

$L_{BC}}$ = 2.5 in

$J\tau$ = 0.098218

G = 11 × 10⁶ lb/in²

$T_{BC$ = -60 lb.ft

$T_{CD$ = 0 lb.ft

$L_{CD$ = 5.5 in

$\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}$

$\phi_{B/D} = \dfrac{-21600}{1079980}$

$\phi_{B/D} =$ − 0.02 rad

To degree; we have

$\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{B/D} = -1.15^0}$

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For the angle of twist of C with respect to D; we have:

$\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{C/D} = \dfrac{21600}{1079980}$

$\phi_{C/D} =$ 0.02 rad

To degree; we have

$\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{C/D} = 1.15^0}$

3 0

3 years ago

H. Blasius correlated data on turbulent friction factor in smooth pipes. His equation f s m o o t h ≈ 0.3164 Re − 1 / 4 fsmooth≈

tiny-mole [99]

Answer:

Therefore the angle the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

Explanation:

The first step to take is to calculate the the velocity of flow through a pipe

Q =Av

Where Q = is the discharge through pipe

A = Area of the pipe

v = the flow of velocity

We substitute 0.001 m^3/s for Q and 0.03 m for D

Q= Av

0.001=Av

Substitute π/4 D² for A

0.001 = π/4 D² (v)

v = 0.004/πD²

D = he diameter of the pipe

substitute 3 cm for D

v= 0.004/π * [3 cm * 1 m/100 cm]²

v =1.414 m/s

Obtain fluid properties from the table Kinematic viscosity and Dynamic of water

p =1000 kg /m³

μ= 1.002 * 10^ ⁻³ N.s/m³

Thus,

we write the expression to determine the Reynolds number of flow

Re = pvD/μ

Re = is the Reynolds number

p =density

μ = dynamic viscosity at 20⁰C

We then substitute 1000 kg /m³ in place of p, 1.002 * 10^ ⁻³ N.s/m³ for μ,

1.414 m/s for v and 0.03 m for D

Thus,

Re = 1000 * 1.414 * 0.03/ 1.002 * 10^ ⁻³ = 42335

The next step is to calculate the friction factor form the Blasius equation

f = 0.3164 (Re)^1/4

f = friction factor

We substitute 42335 for Re

f = 0.3164 (42335)1/4

=0.022

The next step is to write the expression to determine the friction head loss

hl = flv²/2gD

hl = head loss

l = length of pipe

g= acceleration due to gravity

We then again substitute 0.022 for f, 1.414 m/s for v, 0.03 m for D, and 9.8 m/s² for g.

so,

hl = flv²/2gD

hl/L = 0.022 * 1.414²/2 * 9.81 * 0.03

sinθ = 0.07473

θ = 4° 16'

Therefore the angle the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

3 0

3 years ago

A 60-kg woman holds a 9-kg package as she stands within an elevator which briefly accelerates upward at a rate of g/4. Determine

Anuta_ua [19.1K]

Answer:

force R = 846.11 N

lifting force L = 110.36 N

if cable fail complete both R and L will be zero

Explanation:

given data

mass woman mw = 60 kg

mass package mp = 9 kg

accelerates rate a = g/4

to find out

force R and lifting force L and if cable fail than what values would R and L acquire

solution

we calculate here first reaction R force

we know elevator which accelerates upward

so now by direction of motion , balance the force that is express as

R - ( mw + mp ) × g = ( mw + mp ) × a

here put all these value and a = g/4 and use g = 9.81 m/s²

R - ( 60 + 9 ) × 9.81 = ( 60 + 9 ) × g/4

R = ( 69 ) × 9.81/4 + ( 69 ) 9.81

R = 69 ( 9.81 + 2.4525 )

force R = 846.11 N

and

lifting force is express as here

lifting force = mp ( g + a)

put here value

lifting force = 9 ( 9.81 + 9.81/4)

lifting force L = 110.36 N

and

we know if cable completely fail than body move free fall and experience no force

so both R and L will be zero

5 0

3 years ago

In some states, repair shops can accept used oil from do-it-yourselfers or ____________.

MrMuchimi

Answer:

C. Homeowners

Explanation:

I just took my automotive pollution prevention certification and passed by answering homeowners for this question.

7 0

3 years ago

Before installing head gaskets look for “____” labels on the head gaskets

Strike441 [17]

Before installing head gaskets, you should look for "this side up" or "front" labels on the head gaskets.

<h3>What is a head gasket?</h3>

A head gasket can be defined as a gasket which is fitted between the engine block and the cylinder head in an internal combustion engine, so as to seal oil passages and absorb the pressures of the combustion that occurs inside the engine.

As a general rule, you should look for "this side up" or "front" labels on the head gaskets before installing head gaskets in an internal combustion engine of a vehicle.

Read more on head gaskets here: brainly.com/question/1264437

#SPJ1

7 0

2 years ago