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3 years ago

What is the mass in both slugs and kilograms of a 1000-lb beam?

Engineering

1 answer:

Likurg_2 [28]3 years ago

7 0

Answer:

31.1 slug, 453.4 Kg

Explanation:

given,

mass of the beam is 1000 lb

to convert mass of beam into slugs and kilograms.

1 lb is equal to 0.0311 slug

1000 lb = 1000 × 0.0311

= 31.1 slug

now, conversion of lb into kg

1 lb is equal to 0.4534 kg

so,

1000 lb = 1000 × 0.4534

= 453.4 Kg

hence, 1000 lb of beam in slugs is equal to 31.1 slugs and in kilo gram is 453.4 Kg.

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g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th

xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

$qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0$

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }$

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }$

if Gl≈El(l,5800)

$\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt= 2*5800=11600 um-K, at this value, F=0.941

$\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77$

The hemispherical emissivity is equal to:

$E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt=2*333=666 K, at this value, F=0

$E=0+(1-0.7)(1)=0.3$

The hemispherical absorptivity is equal to:

$qe=EoTs^{4}+h(Ts-T\alpha )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}$

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3 years ago

A divided multilane highway in a recreational area has four lanes (two lanes in each direction) and is on rolling terrain. The h

sertanlavr [38]

Answer:

Explanation:

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fHV= 0.847 (Eq. 6.5) PHF = 0.95, fp= 0.90, N=2, V = 2200 (given)

vp= V/[PHF⋅fHVTB⋅fp⋅N] = 1518.9 (Eq. 6.3)

BFFS = 50+5, BFFS =55 (given) fLW= 6.6

TLC=6+3=9 fLC= 0.65

fM= 0.0

fA= 1.0

FFS = BFFS −fLW−fLC–fM–fA= 46.75 (Eq. 6.7)

Use FFS=45 D= vp/S = 33.75pc/mi/ln Eq (6.6)

After: fA= 3.0

FFS = BFFS −fLW−fLC–fM–fA= 44.75 (Eq. 6.7)

Use FFS=45 Vnew= 2600 Vp= Va/[PHF⋅fHB⋅fp⋅N] = 1795 (Eq. 6.3) D= vp/S = 39.89pc/mi/ln

8 0

3 years ago

In a steam power plant, 1 MW is added in the boiler, 0.58 MW is taken out in the condenser and the pump work is 0.02 MW.

Free_Kalibri [48]

Answer:

a) $\eta = 42\,\%$ , b) $COP_{R} = 29$

Explanation:

a) The thermal efficiency is:

$\eta = \frac{\dot Q_{in} - \dot Q_{out}}{\dot Q_{in}}\times 100\,\%$

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$\eta = 42\,\%$

b) The coefficient of performance is:

$COP_{R} = \frac{\dot Q_{L}}{\dot W}$

$COP_{R} = \frac{0.58\,MW}{0.02\,MW}$

$COP_{R} = 29$

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3 years ago

In a planetary geartrain with a form factor of 8, the sun gear rotates clockwise at 5 rad⁄s and the ring gear rotates clockwise

lina2011 [118]

Answer:

D. N= 11. 22 rad/s (CW)

Explanation:

Given that

Form factor R = 8

Speed of sun gear = 5 rad/s (CW)

Speed of ring gear = 12 rad/s (CW)

Lets take speed of carrier gear is N

From Algebraic method ,the relationship between speed and form factor given as follows

$\dfrac{N_{sun}-N}{N_{ring}-N}=-R$

here negative sign means that ring and sun gear rotates in opposite direction

Lets take CW as positive and ACW as negative.

Now by putting the values

$\dfrac{N_{sun}-N}{N_{ring}-N}=-R$

$\dfrac{5-N}{12-N}=-8$

N= 11. 22 rad/s (CW)

So the speed of carrier gear is 11.22 rad/s clockwise.

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3 years ago

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Answer:

a) There is no any private member of smart which are public members of superSmart.

Explanation:

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3 years ago