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4 years ago

can someone help me with this engineering mechanics homework, please? I tried to solve it, but I got so confused.

Engineering

1 answer:

marishachu [46]4 years ago

6 0

Explanation:

Sum of forces in the x direction:

∑Fx = ma

Rx − 250 N = 0

Rx = 250 N

Sum of forces in the y direction:

∑Fy = ma

Ry − 120 N − 300 N = 0

Ry = 420 N

Sum of forces in the z direction:

∑Fz = ma

Rz − 50 N = 0

Rz = 50 N

Sum of moments about the x axis:

∑τx = Iα

Mx + (-50 N)(0.2 m) + (-120 N)(0.1 m) = 0

Mx = 22 Nm

Sum of moments about the y axis:

∑τy = Iα

My = 0 Nm

Sum of moments about the z axis:

∑τz = Iα

Mz + (250 N)(0.2 m) + (-120 N)(0.16 m) = 0

Mz = -30.8 Nm

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3. Assertive communication means:

natta225 [31]

Answer:

A. Confidently and directly communicating what you think and feel, without being aggressive or passive.

Explanation:

Assertive communication is the ability to express both positive and negative ideas in a way that it is a honest and direct message. It doesn't mean to say things to the other person with the intent of causing harm or making them feel better, it's just saying things as they are.

7 0

2 years ago

A tensile test was made on a tensile specimen, with a cylindrical gage section which had a diameter of 10 mm, and a length of 40

tamaranim1 [39]

Answer:

The answers are as follow:

a) 10 mm

b) 12.730 N/ $mm^{2}$

c) 127.307 N/ $mm^{2}$

d) 0.25

Explanation:

d1 = 10mm , L1 = 40 mm, L2 = 50 mm, reduction in area = 90% = 0.9

Force = F =1000 N

let us find initial area first, A1 = pi* $r^{2}$ = 78.55 $mm^{2}$

using reduction in area formula : 0.9 = (A1 - A2 ) / A1

solving it will give, A2 = 0.1 A1 = 7.855 $mm^{2}$

a) The specimen elongation is final length - initial length

50 - 40 = 10 mm

b) Engineering stress uses the original area for all stress calculations,

Engineering stress = force / original area = F / A1 = 1000 / 78.55

Engineering stress = 12.730 $N / mm^{2}$

c) True stress uses instantaneous area during stress calculations,

True fracture stress = force / final area = F / A2 = 1000 / 7.855

True Fracture stress = 127.30 $N / mm^{2}$

e) strain = change in length / original length

strain = 10 / 40 = 0.25

8 0

3 years ago

A 12-ft high retaining wall has backfill of granular soil with an internal angle of friction of 30 and unit weight of 125 pef. W

irakobra [83]

Answer:

P_p = 27000 psf

Explanation:

given,

height of the retaining wall = h = 12 ft

internal angle of friction (∅)= 30°

unit weight = 125 pcf

Rankine passive earth pressure = ?

k_p is the coefficient of passive earth pressure

$k_p = \dfrac{1 + sin\phi}{1 - sin\phi}$

$k_p = \dfrac{1 + sin30^0}{1 - sin30^0}$

k_p = 3

Passive earth pressure

$P_p = \dfrac{1}{2}k_p \gamma H^2$

$P_p = \dfrac{1}{2}\times 3\times 125 \times 12^2$

P_p = 27000 psf

Rankine passive earth pressure on the wall is equal to P_p = 27000 psf

7 0

3 years ago

A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is re

Anna [14]

Answer:

A. 72.34mol/min

B. 76.0%

Explanation:

We start by converting to molar flow rate. Using density and molecular weight of hexane

= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17

= 988.5/86.17

= 11.47mol/min

n1 = n2+n3

n1 = n2 + 11.47mol/min

We have a balance on hexane

n1y1C6H14 = n2y2C6H14 + n3y3C6H14

n1(0.18) = n2(0.05) + 11.47(1.00)

To get n2

(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)

0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min

0.18n2-0.05n2 = 11.47-2.0646

= 0.13n2 = 9.4054

n2 = 9.4054/0.13

n2 = 72.34 mol/min

This value is the flow rate of gas that is leaving the system.

n1 = n2 + 11.47mol/min

72.34mol/min + 11.47mol/min

= 83.81 mol/min

Amount of hexane entering condenser

0.18(83.81)

= 15.1 mol/min

Then the percentage condensed =

11.47/15.1

= 7.59

~7.6

7.6x100

= 76.0%

Therefore the answers are a.) 72.34mol/min b.) 76.0%

Please refer to the attachment .

4 0

3 years ago

Construct a Mohr circle for the stress element at A in problem 2. Using ruler and compass, draw the Mohr circle to the scale. Dr

serious [3.7K]

Answer:

hello your question is incomplete attached below is the missing diagram to the question and the detailed solution

Answer : principal stresses : 0.82 MPa, -33.492 MPa

shear stress = 17.157 MPa

∅ = 9.09 ≈ 10°

Explanation:

The principal stress ( б1 ) = 0.82 MPa

( б2 ) = -33.492 MPa

The shear stress = 17.157 MPa

∅ = 9.09 ≈ 10°

attached below is the detailed solution and the Mohr's circle

7 0

3 years ago

can someone help me with this engineering mechanics homework, please? I tried to solve it, but I got so confused.​

can someone help me with this engineering mechanics homework, please? I tried to solve it, but I got so confused.