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denis23 [38]
4 years ago
14

Calculate the length of a metal cylinder while it is subjected to a tensile stress of 10,000 psi. You are given the following da

ta: Original length = 1 in Original cross-sectional area = 0.1 in2 Yield strength, σy = 9 ksi Young’s modulus, E = 1000 ksi
Engineering
1 answer:
anastassius [24]4 years ago
5 0

Answer:

length of cylinder can not calculated

Explanation:

given data

tensile stress = 10,000 psi

Original length = 1

Original cross-sectional area = 0.1 in²

Yield strength, σy = 9 ksi

Young’s modulus, E = 1000 ksi

solution

we can see that here that applied stress is greater than yield stress of material  that is express

1000 ksi  >  9 ksi

so here hooks law and strain relation is not working

so length of cylinder can not calculated

as stress applied 10000 psi

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I think the answer is B. Black

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A slug travels 3 centimeters in 3 seconds. A snail travels 6 centimeters in 6 seconds. Both travel at constant speeds. Mai says,
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i dont agree with mai because they were both going 1cm per second

Explanation:

3÷3=1

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3 years ago
Select three functions of catalysts.
Korolek [52]

Answer: speed up food processing

speed up plant growth

Increase fuel efficiency

Explanation:

A catalyst simply refers to a substance that leads to an increase in the reaction rate when it's added to a substance. When the activation energy is reduced by catalysts, this.hwlpa on the speeding up of a reaction.

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3 years ago
A helicopter moves horizontally in the x direction at a speed of 120 mi/h. Knowing that the main blades rotate clockwise when vi
devlian [24]

Answer:

The instantaneous axis of rotation=

x = 0 ; z = 8.4 ft

Explanation:

Given:

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\frac{5280 * 120}{60*60}

= 176 ft/s

Angular velociyy, w = 220 rpm, converting to rad/sec, we have: \frac{200*2*pi}{60} =20.95 rad/s

The helicopter moves horizontally in the x direction at a speed of 120 mi/h, this means that the helicopter moves in the positive x direction at 120mi/h

To find the instantaneous axis of rotation of the main blades, we have:

Where Vc = 20.95 rad/s

Vo = 176 ft/s

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= 8.4 ft

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4 0
3 years ago
The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp
Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1

The Lower Surface Cp is given as

Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1

The difference of the Cp over the airfoil is given as

\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32

Now the Lift Coefficient is given as

C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192

Now the coefficient of moment about the leading edge is given as

C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0
3 years ago
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