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2 years ago

Select the statement that is false.

Engineering

1 answer:

ra1l [238]2 years ago

8 0

Answer:

Explanation:

the way vertices are connected may be different so having same number of edges do not mean that total degree will also be same.

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A cast-iron tube is used to support a compressive load. Knowing that E 5 10 3 106 psi and that the maximum allowable change in l

Papessa [141]

Answer:

(a) 2.5 ksi

(b) 0.1075 in

Explanation:

(a)

$E=\frac {\sigma}{\epsilon}$

Making $\sigma$ the subject then

$\sigma=E\epsilon$

where $\sigma$ is the stress and $\epsilon$ is the strain

Since strain is given as 0.025% of the length then strain is $\frac {0.025}{100}=0.00025$

Now substituting E for $10\times 10^{6} psi$ then

$\sigma=(10\times 10^{6} psi)\times 0.00025=2500 si= 2.5 ksi$

(b)

Stress, $\sigma= \frac {F}{A}$ making A the subject then

$A=\frac {F}{\sigma}$

$A=\frac {\pi(d_o^{2}-d_i^{2})}{4}$

where d is the diameter and subscripts o and i denote outer and inner respectively.

We know that $2t=d_o - d_i$ where t is thickness

Now substituting

$\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}$

$\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4$

$(d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

But the outer diameter is given as 2 in hence

$(2^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

$2^{2}-(\frac {1600}{2500\times \pi}\times 4)=d_i^{2}$

$d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 4)}=1.784692324 in\approx 1.785 in$

As already mentioned, $2t=d_o - d_i hence t=0.5(d_o - d_i)$

$t=0.5(2-1.785)=0.1075 in$

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3 years ago

O local utilizado pelos grandes avioes para descolar e aterrar

IRISSAK [1]

Decolagem e pouso

espero que ajude

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2 years ago

Print reading for industry unit 9 review questions

RideAnS [48]

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3 years ago

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Subcooled liquid water flows adiabatically in a constant diameter pipe past a throttling valve that is partially open. The liqui

Llana [10]

Answer:

hi-he = 0

pi-pe = positive

ui-ue = negative

ti-te = negative

Explanation:

we know that fir the sub cool liquid water is

dQ = Tds = du + pdv ............1

and Tds = dh - v dP .............2

so now for process of throhling is irreversible when v is constant

then heat transfer is = 0 in irreversible process

so ds > 0

so here by equation 1 we can say

ds > 0

dv = 0 as v is constant

so that Tds = du .................3

and du > 0

ue - ui > 0

and

now by the equation 2 throttling process

here enthalpy is constant

so dh = 0

and Tds = -vdP

so ds > 0

so that -vdP > 0

as here v is constant

so -dP =P1- P2

so P1-P2 > 0

so pressure is decrease here

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3 years ago

Hi im ***ar and im doing sculptural but what should it be about star wars or Marvel

Anarel [89]

Answer:

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3 years ago