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2 years ago

A master stud pattern is laid out somewhat like a?

Engineering

1 answer:

Svetradugi [14.3K]2 years ago

3 0

Answer:

••• like a story pole but has information for only one portion of the wall. system. methods and materials of construction.

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5. The pin support at A allows _______. Select the one that applies. (a) displacement in the x direction (b) rotation about its

soldier1979 [14.2K]

Answer: Diagram associated with your question is attached below

5) B

6) C

Explanation:

5) The pin support at A allows ; Rotation about its central axis

This is because pin supports does not allow the translation of its structural member in any direction i.e. y or x but only rotation about its axis

6) The support at B does not allow displacement in y direction

This is because roller support allows displacement only in the direction that they are situated and in this case it is the x - direction

7 0

3 years ago

What is the relationship between orifice diameter and pipe diameter

NeX [460]

Answer:

......................................................

4 0

3 years ago

If a motorist moves with a speed of 30 km/hr, and covers the distance from place A to place B

Sergio039 [100]

Answer:

105 km

Explanation:

The motorist was going 30 km/hr, and it took 3 hours 30 minutes. That's 3.5 hours. 3.5×30=105

5 0

4 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

3 years ago

Can you solve this question

Alecsey [184]

Answer:

eojcjksjsososisjsiisisiiaodbjspbcpjsphcpjajosjjs ahahhahahahahahahahahahahahahhhahahahaahahhahahahahaahahahahaha

6 0

3 years ago

Read 2 more answers