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3 years ago

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A 240-ton tugboat is moving at 6 ft/s with a slack towing cable attached to a 100-ton barge that is at rest. The cable is being

unwound from a drum on the tugboat at a constant rate of 5.4 ft/s and that rate is maintained after the cable becomes taut.
The velocity of the tugboat after the cable becomes taut is:_________

1 answer:

Zolol [24]3 years ago

5 0

Nobody about to do all of that, lol

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A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is 30 cm in diame

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Answer:

Q = 125.538 W

Explanation:

Given data:

D = 30 cm

Temperature $T_\infity = 15$ degree celcius

$T_S = 220 + 273 = 473 K$

Heat coefficient = 12 W/m^2 K

Efficiency 80% = 0.8

$Q = hA(T_S - T_{\infty}) \eta$

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Engine oil flows at a rate of 1 kg/s through a 5-mmdiameter straight tube. The oil has an inlet temperature of 45°C and it is de

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Answer:

length of tube = 4.12 m

Explanation:

Given data:

flow rate of engine oil = 1 kg/s

diameter of tube = 5-mm

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exit temperature of oil = 80°C

surface temperature of tube = 150°C

<u>Determine the required length of the tube </u>

attached below is a detailed solution to the given problem

length of tube = 4.12 m

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3 years ago

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stre

velikii [3]

Complete Question

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.

Answer:

The stress is $\sigma = 10. 655 MPa$

Explanation:

From the question we are told that

The critical yield resolved shear stress is $\sigma = 2.9Mpa$

First we obtain the angle $\lambda$ between the slip direction [121] and [111]

$\lambda = cos^{-1} [\frac{(u_1 u_2 + v_1 v_2 + w_1 w_2}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_2^2 + v_2^2 + w_2 ^2)} } ]$

Where $u_1 ,u_2 ,v_1 , v_2 , w_1 , w_2$ are the directional indices

$\lambda = cos ^-[ \frac{(1) (-1) + (2) (1) + (1) (1)}{\sqrt{((1)^2 +(2)^2 + (1)^2)}\sqrt{((-1)^2 + (1)^2 + (1)^2 ) } } ]$

$= cos^{-1} [\frac{2}{\sqrt{6} \sqrt{3} } ]$

$= 61.87^0$

Next is to obtain the angle $\O$ between the direction [121] and [101]

$\O = cos^{-1} [\frac{(u_1 u_3 + v_1 v_3 + w_1 w_3}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_3^2 + v_3^2 + w_3 ^2)} } ]$

Substituting 1 for $u_1$ , 2 for $v_1$ , 1 for $w_1$ , 1 for $u_2$ , 0 for $v_2$ , and 1 for $w_2$

$\O = cos^{-1} [\frac{1* 1 + 2*0 + 1*1 }{\sqrt{1^2 + 2^2 + 1^2 } \sqrt{(1^2 + 0^2 + 1^2 )} } ]$

$\O = cos^{-1} [\frac{2}{\sqrt{6} \sqrt{2} } ]$

$= 54.74 ^o$

The stress is mathematically represented as

$\sigma = \frac{\tau_c}{cos \O cos \lambda }$

$= \frac{2.9}{cos 54.74^o cos 61.87^o}$

$= \frac{2.9}{0.2722}$

$\sigma = 10. 655 MPa$

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