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4 years ago

A cable in a motor hoist must lift a 700-lb engine. The steel cable is 0.375in. in diameter. What is the stress in the cable?

Engineering

1 answer:

UkoKoshka [18]4 years ago

7 0

Answer:43.70 MPa

Explanation:

Given

mass of engine $700 lb \approx 317.515 kg$

diameter of cable $0.375 in.\approx 9.525 mm$

$A=\frac{\pi d^2}{4}=71.26 mm^2$

we know stress( $\sigma$ ) $=\frac{load\ applied}{area\ of\ cross-section}$

$\sigma =\frac{317.515\times 9.81}{71.26\times 10^{-6}}=43.70 MPa$

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how to take care of a kid?? this kid is begging me to do something i dont want to help please its a 5yo

Finger [1]

Answer: Feed him, give him meds

Explanation:

8 0

3 years ago

5) Calculate the LMC wal thickness of a pipe and tubing with OD as 35 + .05 and ID as 25 + .05 A) 4.95 B) 5.05 C) 10 D) 15.025

Radda [10]

Answer:

LMC wall thickness= 5.05

Explanation:

Given:

Minimum inner diameter = 25 - 0.05 = 24.95

Maximum outer diameter = 35 + 0.05 = 35.05

Find:

LMC wall thickness

Computation:

LMC wall thickness = (maximum outer diameter - minimum inner diameter) / 2

LMC wall thickness = (35.05 - 24.95) / 2

LMC wall thickness= 5.05

3 0

3 years ago

Consider a single crystal oriented such that the slip direction and normal to the slip plane are at angles 42.7° and 48.3°, resp

zlopas [31]

Answer:

Applied Stress > 58.29 MPa

Explanation:

Resolved shear stress should be greater than critically resolved shear stress in order to cause the single crystal to yield

Given angles are

∅ = 42.7 degree

Ф = 48.3 degree

Critically resolved shear stress = 28.5 MPa

If we consider

Critically resolved shear stress = resolved shear stress

Applied stress can be found by

$Z_{R} = applied stress X cos\phi X cos\theta$ (1)

Applied Stress = $\frac{Z_{R} }{Cos\phi XCos\theta}$

Applied Stress = $\frac{28.5}{Cos(48.3)XCos(42.7)}$

Applied Stress = 58.29 MPa

We got reference

By putting applied stress values of greater than 58.29 MPa in equation 1 we get

Resolved Shear Stress = 60 x Cos(48.3) x Cos(42.7)

Resolved Shear Stress = 29.33 MPa

Therefore, by the above calculation we conclude that applied stress should be greater than 58.29 MPa, In order to make resolved shear stress to be greater than critically resolved shear stress that is essential for single crystal to yield.

8 0

3 years ago

An incompressible viscous fluid flows through a pipe with a flow rate of 1 mL/s. The pipe has a uniform diameter D0 and a length

Nataly [62]

Answer:

$Q_2 = 32$ mL/s

Explanation:

Given :

The flow is incompressible viscous flow.

The initial flow rate, $Q_1$ = 1 mL/s

Initial diameter, $D_1= D_0$

Initial length, $L_1=L_0$

The initial pressure difference to maintain the flow, $P_1=P_0$

We know for a viscous flow,

$\Delta P = \frac{32 \mu V L}{D^2}$

$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$

$Q \propto \Delta P \times D^4$

$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$

$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$

$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$

$\frac{1}{Q_2}= \frac{1}{32}$

∴ $Q_2 = 32$ mL/s

6 0

3 years ago

The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of t

Tju [1.3M]

Answer:

15.99 ft/s

Explanation:

From Newton's equation of motion, we have

v = u + at

v = Final speed

u = initial speed

a = acceleration

t = time

now

for the points A and C

v = 17.6 ft/s

u = 13.2 ft/s

t = 3 s

thus,

17.6 = 13.2 + a(3)

3a = 17.6 - 13.2

3a = 4.4

a = 1.467 m/s²

Thus,

For Points A and B

v = speed at B i.e v'

u = 13.2 ft/s

a = 1.467 ft/s²

t = 1.90 s

therefore,

v' = 13.2 + (1.467 × 1.90 )

v' = 13.2 + 2.7867

v' = 15.9867 ≈ 15.99 ft/s

3 0

3 years ago