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3 years ago

This might count as engineering, I'm not sure as this is IT

Engineering

1 answer:

trasher [3.6K]3 years ago

6 0

Answer:

10.5

Explanation:

Convert to an equation for example P%* X=Y

P is 7.5% X is 140, so the equation Is 7.5 percent * 14= Y

convert 7.5% Into a decimal by removing the percent sign and deviding by 7.5/100= 0.075

Substitute 0.075 for 7.5% in the equation: 7.5%*140=Y becomes 0.075*140= 10.5

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48/64 reduced to its lowest term

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Answer:

3/4

Explanation:

48/64 = 3/4

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3 years ago

Read 2 more answers

The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 2300 lb⋅ft, a

Arturiano [62]

Answer:

Explanation:

Given that:

Torque T = 2300 lb - ft

Bending moment M = 1500 lb - ft

axial thrust P = 2500 lb

yield points for tension σY= 100 ksi

yield points for shear τY = 50 ksi

Using maximum-shear-stress theory

$\sigma_A = \dfrac{P}{A}+\dfrac{Mc}{I}$

where;

$A = \pi c^2$

$I = \dfrac{\pi}{4}c^4$

$\sigma_A = \dfrac{P}{\pi c^2}+\dfrac{Mc}{ \dfrac{\pi}{4}c^4}$

$\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{1500*12c}{ \dfrac{\pi}{4}c^4}$

$\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{72000c}{\pi c^3}}$

$\tau_A = \dfrac{T_c}{\tau}$

where;

$\tau = \dfrac{\pi c^4}{2}$

$\tau_A = \dfrac{T_c}{\dfrac{\pi c^4}{2}}$

$\tau_A = \dfrac{2300*12 c}{\dfrac{\pi c^4}{2}}$

$\tau_A = \dfrac{55200 }{\pi c^3}}$

$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\dfrac{(\sigma_x - \sigma_y)^2}{2}+ \tau_y^2}$

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi c ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi c^3}+ \dfrac{55200}{\pi c^3}} \ \ \ \ \ ------(1)$

Let say :

$|\sigma_1 - \sigma_2| = \sigma_y$

Then :

$2\sqrt{( \dfrac{2500c + 72000}{2 \pi c^3})^2+ ( \dfrac{55200}{\pi c^3})^2 } = 100(10^3)$

$(2500 c + 72000)^2 +(110400)^2 = 10000*10^6 \pi^2 c^6$

$6.25c^2 + 360c+ 17372.16-10,000\ \pi^2 c^6 =0$

According to trial and error;

c = 0.75057 in

Replacing c into equation (1)

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}$

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} + \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}} \ \ \ OR \\ \\ \\ \sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} - \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}$

$\sigma _1 = 22193 \ Psi$

$\sigma_2 = -77807 \ Psi$

The required diameter d = 2c

d = 1.50 in or 0.125 ft

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Answer:

Explanation:

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nevsk [136]

Answer:

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3 years ago

Design a program that calculates the area and circumstance of rectangle?

Phantasy [73]

Let “w” and “L” be the width and length of the rectangle. “p” and “a” are perimeter and area
For python,
w=int(input(“width”))
l=int(input(“length”))
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p=2*w+2*l
print(str(a), str(p)

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2 years ago