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brilliants [131]

3 years ago

13

*6–24. The beam is used to support a dead load of 400 lb>ft, a live load of 2 k>ft, and a concentrated live load of 8 k. D

etermine (a) the maximum positive vertical reaction at A, (b) the maximum positive shear just to the right of the support at A, and (c) the maximum negative moment at C. Assume A is a roller, C is fixed, and B is pinned.

1 answer:

lisabon 2012 [21]3 years ago

3 0

Answer:

(a) maximum positive reaction at A = 64.0 k

(b) maximum positive shear at A = 32.0 k

(c) maximum negative moment at C = -540 k·ft

Explanation:

Given;

dead load Gk = 400 lb/ft

live load Qk = 2 k/ft

concentrated live load Pk =8 k

(a) from the influence line for vertical reaction at A, the maximum positive reaction is

$A_{ymax}$ = 2*(8) +(1/2(20 - 0)* (2))*(2 + 0.4) = 64 k

See attachment for the calculations of (b) & (c) including the influence line

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tatuchka [14]

Answer:

Check the v(t) signal referred to in the question and the solution to each part in the files attached

Explanation:

The detailed solutions of parts a to d are clearly expressed in the second file attached.

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3 years ago

Plan to refuel when your fuel gauge reads __________.

bazaltina [42]

Answer:

low on fuel or if it's red

Explanation:

common sense to be honest :/

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2 years ago

Read 2 more answers

A heating system must maintain the interior of a building at 20°C during a period when the outside air temperature is 5°C and th

Anettt [7]

Answer:

a. W = 51,194.54 kJ

b. W = 102,390 kJ

c. W = 153,585 kJ

Explanation:

$(COP)_{HP} =\frac{Desired-effectx}{Work-done}= \frac{Q_{1} }{W} \\\\(COP)_{HP} =(COP)_{Ideal}\\\\\frac{Q1}{W} =\frac{T_{1} }{T_{1} -T_{2} }$

$W=Q_{1} \frac{T_{1}-T_{2} }{T_{1} }$

a. the ground at 15°C.

$T_{1}$ =20°C = 273 K + 20 = 293 K

$T_{2}$ =15°C = 273 K + 15 = 288 K

$Q_{1}=3x10^{6} kJ$

$W=3x10^{6} kJ \frac{293 K-288 K}{293 K}=3x10^{6} kJ \frac{5 K}{293 K}=3x10^{6} kJ x 0.017065}$

$W = 0.051195x10^{6} kJ$

W = 51,194.54 kJ

b. a pond at 10°C.

$T_{2}$ =10°C = 273 K + 10 = 283 K

$W=3x10^{6} kJ \frac{293 K-283 K}{293 K}=3x10^{6} kJ \frac{10 K}{293 K}=3x10^{6} kJ x 0.034130}$

$W = 0.102390x10^{6} kJ$

W = 102,390 kJ

c. the outside air at 5°C.

$T_{2}$ =5°C = 273 K + 5 = 278 K

$W=3x10^{6} kJ \frac{293 K-278 K}{293 K}=3x10^{6} kJ \frac{15 K}{293 K}=3x10^{6} kJ x 0.051195}$

$W = 0.153585x10^{6} kJ$

W = 153,585 kJ

Hope this helps!

3 0

2 years ago

A hypothetical metal has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.413 nm, 0.665 nm, and 0.87

Inessa [10]

Answer:

atomic radius R = 0.157 nm

metal atomic weight = 72.27 g/mol

Explanation:

given data

parameters a = 0.413 nm

parameters b = 0.665 nm

parameters c = 0.876 nm

atomic packing factor = 0.536

density = 3.99 g/cm³

to find out

atomic radius and atomic weight

solution

we apply here atomic packing factor (x) that is

atomic packing factor (x) = $\frac{volume(sphere)}{volume(unit\ cell)}$ ..................1

put here value we get

atomic packing factor = $\frac{8*(4/3)*\pi R^3}{3*a*b*c}$

R = $(\frac{3(x)(abc)}{32\pi })^{1/3}$

R = $(\frac{3(0.536)(0.413*0.665*0.876)}{32\pi })^{1/3}$

atomic radius R = 0.157 nm

and

now we get here metal atomic weight that is

metal atomic weight = $\frac{\rho (abc)(N_A)}{no\ of\ atom}$ ....................2

metal atomic weight = $\frac{3.99 (0.413*0.665*0.876)(6.023*10^{23})}{8}$

metal atomic weight = 72.27 g/mol

6 0

3 years ago

The cross section of a heat exchanger consists of three circular pipes inside a larger pipe. The internal diameter of the three

omeli [17]

Answer:

0.0432 m^3/s

Explanation:

Internal diameter of smaller pipes = 2.5 cm = 0.025 m

pipa wall thickness = 3 mm = 0.003 m

internal diameter of larger pipes = 8 cm = 0.8 m

velocity of region between smaller and larger pipe = 10 m/s

Calculate discharge in m^3/s

First we calculate the area of the smaller pipe

A = $\pi Dt$ = $\pi ( 0.025 ) ( 0.003 )$ = 0.00023571 m^2

next we calculate area of fluid between the smaller pipes and larger pipe

A = $[\frac{\pi }{4} D^{2} _{L} ] - 3(A_{s})$

= $[ \frac{\pi }{4} (0.08 )^2 - 3 ( 0.00023571 )]$

= [ 0.00502857 - 0.00070713 ]

= 0.00432144 m^2

hence the discharge in m^3/s

Q = AV

= 0.00432144 * 10

= 0.0432 m^3/s

3 0

2 years ago