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3 years ago

Is there a way to get the answers to a NCCER book test?

Engineering

1 answer:

sergeinik [125]3 years ago

3 0

Answer:

go on google and type NCEER book answers

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A circular specimen of MgO is loaded in three-point bending. Calculate the minimum possible radius of the specimen without fract

Hitman42 [59]

Answer:

radius = 9.1 × $10^{-3}$ m

Explanation:

given data

applied load = 5560 N

flexural strength = 105 MPa

separation between the support = 45 mm

solution

we apply here minimum radius formula that is

radius = $\sqrt[3]{\frac{FL}{\sigma \pi}}$ .................1

here F is applied load and is length

put here value and we get

radius = $\sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}$

solve it we get

radius = 9.1 × $10^{-3}$ m

8 0

3 years ago

After reading through the code, what will happen when you click run?* 1 point when run move forward while there is a pile do rem

KATRIN_1 [288]

Explanation:

I'm not exactly a master at coding, but I'm pretty sure that:

The farmer will remove dirt as long as there is a pile, then stop when the pile is done.

5 0

2 years ago

Hęłłõ hõw årę ÿõū dõìńg tõdāÿ

kirill [66]

Answer:

All good :)

What about you??

8 0

2 years ago

Read 2 more answers

An incompressible fluid flows between two infinite stationary parallel plates. The velocity profile is given by u=umaxðAy2 + By+

nexus9112 [7]

Answer:

the volume flow rate per unit depth is:

$\frac{Q}{b} = \frac{2}{3} u_{max} h$

the ratio is : $\frac{V}{u_{max}}=\frac{2}{3}$

Explanation:

From the question; the equations of the velocities profile in the system are:

$u = u_{max}(Ay^2+By+C)$ ----- equation (1)

The above boundary condition can now be written as :

At y= 0; u =0 ----- (a)

At y = h; u =0 -----(b)

At y = $\frac{h}{2}$ ; u = $u_{max}$ ------(c)

where ;

A,B and C are constant

h = distance between two plates

u = velocity

$u_{max}$ = maximum velocity

y = measured distance upward from the lower plate

Replacing the boundary condition in (a) into equation (1) ; we have:

$u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*0+B*0+C) \\ \\ 0=u_{max}C \\ \\ C= 0$

Replacing the boundary condition (b) in equation (1); we have:

$u = u_{max}(Ay^2+By+C) \\ \\ 0 = u_{max}(A*h^2+B*h+C) \\ \\ 0 = Ah^2 +Bh + C \\ \\ 0 = Ah^2 +Bh + 0 \\ \\ Bh = - Ah^2 \\ \\ B = - Ah \ \ \ \ \ --- (d)$

Replacing the boundary condition (c) in equation (1); we have:

$u = u_{max}(Ay^2+By+C) \\ \\ u_{max}= u_{max}(A*(\frac{h^2}{2})+B*\frac{h}{2}+C) \\ \\ 1 = \frac{Ah^2}{4} +B \frac{h}{2} + 0 \\ \\ 1 = \frac{Ah^2}{4} + \frac{h}{2}(-Ah) \\ \\ 1= \frac{Ah^2}{4} - \frac{Ah^2}{2} \\ \\ 1 = \frac{Ah^2 - Ah^2}{4} \\ \\ A = -\frac{4}{h^2}$

replacing $A = -\frac{4}{h^2}$ for A in (d); we get:

$B = - ( -\frac{4}{h^2})h$ $B = \frac{4}{h}$

replacing the values of A, B and C into the velocity profile expression; we have:

$u = u_{max}(Ay^2+By+C) \\ \\ u = u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y)$

To determine the volume flow rate; we have:

$Q = AV \\ \\ Q= \int\limits^h_0 (u.bdy)$

Replacing $u_{max} (-\frac{4}{h^2}y^2+\frac{4}{h}y) \ for \ u$

$\frac{Q}{b} = \int\limits^h_0 u_{max}(-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\ \frac{Q}{b} = u_{max} \int\limits^h_0 (-\frac{4}{h^2} y^2+\frac{4}{h}y)dy \\ \\ \frac{Q}{b} = u_{max} (-\frac{-4}{h^2}\frac{y^3}{3} +\frac{4}{h}\frac{y^2}{y})^ ^ h}}__0 }} \\ \\ \frac{Q}{b} =u_{max} (-\frac{-4}{h^2}\frac{h^3}{3} +\frac{4}{h}\frac{h^2}{y})^ ^ h}}__0 }} \\ \\ \frac{Q}{b} = u_{max}(\frac{-4h}{3}+\frac{4h}2} ) \\ \\ \frac{Q}{b} = u_{max}(\frac{-8h+12h}{6}) \\ \\ \frac{Q}{b} =u_{max}(\frac{4h}{6})$

$\frac{Q}{b} = u_{max}(\frac{2h}{3}) \\ \\ \frac{Q}{b} = \frac{2}{3} u_{max} h$

Thus; the volume flow rate per unit depth is:

$\frac{Q}{b} = \frac{2}{3} u_{max} h$

Consider the discharge ;

Q = VA

where :

A = bh

Q = Vbh

$\frac{Q}{b}= Vh$

Also; $\frac{Q}{b} = \frac{2}{3} u_{max} h$

Then;

$\frac{2}{3} u_{max} h = Vh \\ \\ \frac{V}{u_{max}}=\frac{2}{3}$

Thus; the ratio is : $\frac{V}{u_{max}}=\frac{2}{3}$

5 0

3 years ago

Determine ten different beam loading values that will be used in lab to end load a cantilever beam using weights. Load values sh

nasty-shy [4]

Answer:

1st value = 1.828 * 10 ^9 gm/m^2 ------- 10th value = 7.312 * 10^9 gm/m^2

Explanation:

initial load ( Wp) = 200 g

W1 ( value by which load values increase ) = 100 g

Ten different beam loading values :

Wp + w1 = 300g ----- p1

Wp + 2W1 = 400g ---- p2

Wp + 3W1 = 500g ----- p3 ----------------- Wp + 10W1 = 1200g ---- p10

x = 10.25" = 0.26 m

b = 1.0" = 0.0254 m

t = 0.125" = 3.175 * 10^-3 m

using the following value to determine the load values at different beam loading values

attached below is the remaining part fo the solution

5 0

3 years ago