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victus00 [196]
2 years ago
12

6.39 A particular machine part is subjected in service to a maximum load of 10 kN. With the thought of providing a safety factor

of 1.5, it is designed to withstand a load of 15 kN. If the maximum load encountered in various applications is normally distributed with a standard deviation of 2 kN, and if part strength is normally distributed with a standard deviation of 1.5 kN, what failure percentage would be expected in service? [Ans.: 2.3%]

Engineering
1 answer:
Serggg [28]2 years ago
8 0

Answer:

2.275 %

Explanation:

see the attached picture for detailed answer.

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For goods-producing firms, at which of the following levels of resource planning does scheduling for individual subassemblies an
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Disaggregation

Explanation:

In a company it is a way to create operational plans that are focused, either by time or by section.

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A BOD test is run using 30 mL of wastewater and 270 mL of dilution water. The initial DO of the mixture is 9.0 mg/L. After 5 day
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The clepsydra, or water clock, was a device that the ancient Egyptians, Greeks, Romans, and Chinese used to measure the passage
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Suppose a tank is made of glass and has the shape of a right-circular cylinder of radius 1 ft. Assume that h(0) = 2 ft corresponds to water filled to the top of the tank, a hole in the bottom is circular with radius in., g = 32 ft/s2, and c = 0.6. Use the differential equation in Problem 12 to find the height h(t) of the water.

Answer:

Height of the water = √(128)/147456 ft

Explanation:

Given

Radius, r = 1 ft

Height, h = 2 ft

Radius of hole = 1/32in

Acceleration of gravity, g = 32ft/s²

c = 0.6

Area of the hold = πr²

A = π(1/32)² ---- Convert to feet

A = π(1/32 * 1/12)²

A = π/147456 ft²

Area of water = πr²

A = π 1²

A = π

The differential equation is;

dh/dt = -A1/A2 √2gh where A1 = Area of the hole and A2 = Area of water

A1 = π/147456, A2 = π

dh/dt = (π/147456)/π √(2*32*2)

dh/dt = 1/147456 * √128

dh/dt = √128/147456 ft

Height of the water = √(128)/147456 ft

3 0
3 years ago
A 15,000 kg satellite is orbiting at 800 km above the Earth’s surface. What is the potential energy of the satellite?
vredina [299]

Answer:

-8.34\times 10^{11}\ \text{J}

Explanation:

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h = Distance above Earth = 800 km

R = Radius of Earth = 6371 km

M = Mass of Earth = 5.972\times 10^{24}\ \text{kg}

G = Gravitational constant = 6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2

Potential energy is given by

U=-\dfrac{GMm}{R+h}\\\Rightarrow U=-\dfrac{6.674\times 10^{-11}\times 5.972\times 10^{24}\times 15000}{(6371+800)\times 10^3}\\\Rightarrow U=-8.34\times 10^{11}\ \text{J}

The potential energy of the satellite is -8.34\times 10^{11}\ \text{J}.

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