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3 years ago

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If block A of the pulley system is moving downward at 6 ft>s while block C is moving down at 18 ft>s, determine the relati

ve velocity of block B with respect to C.

1 answer:

kramer3 years ago

3 0

Answer:

Explanation:

The detailed steps and appropriate calculation with analysis is as shown in the attachment.

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3.8 LAB - Select lesson schedule with multiple joins

dem82 [27]

Answer:

The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key; RegisteredName; Breed; Height; BirthDate.

Explanation:

4 0

2 years ago

A company purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate

olga2289 [7]

Answer:

1) The probability of at least 1 defective is approximately 45.621%

2) The probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is approximately 16.0212%

Explanation:

The given parameters are;

The defective rate of the device = 3%

Therefore, the probability that a selected device will be defective, p = 3/100

The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;

The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97

The probability of 0 defective in 20 = ₂₀C₀(0.03)⁰·(0.97)²⁰ ≈ 0.543794342927

The probability of at least 1 = 1 - The probability of 0 defective in 20

∴ The probability of at least 1 = 1 - 0.543794342927 = 0.45621

The probability of at least 1 defective ≈ 0.45621 = 45.621%

2) The probability of at least 1 defective in a shipment, p ≈ 0.45621

Therefore, the probability of not exactly 1 defective = q = 1 - p

∴ q ≈ 1 - 0.45621 = 0.54379

The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;

P(Exactly 3 with at least 1) = ₁₀C₃(0.45621)³(0.54379)⁷ ≈ 0.160212

Therefore, the probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is 16.0212%

4 0

3 years ago

A 4140 steel shaft, heat-treated to a minimum yield strength of 100 ksi, has a diameter of 1 7/16 in. The shaft rotates at 600 r

velikii [3]

Answer:

Explanation:

4140-40 I’d pick wood

I hope this helps! :)

4 0

3 years ago

Read 2 more answers

A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 Mpa root m is exposed to a stress of 1030 MPa. W

cupoosta [38]

Answer:

It will not experience fracture when it is exposed to a stress of 1030 MPa.

Explanation:

Given

Klc = 54.8 MPa √m

a = 0.5 mm = 0.5*10⁻³m

Y = 1.0

This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:

<em>σc = KIc / (Y*√(π*a))</em>

Thus

σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))

⇒ σc = 1382.67 MPa > 1030 MPa

Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.

3 0

3 years ago

A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas

ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × $10^{5}$ Pa

P (gauge ) = 2 × $10^{5}$ Pa

so from idea gas equation

$\frac{P1*V1}{T1} = \frac{P2*V2}{T2}$ ................1

so ${P2} = \frac{P1*T2}{T1}$

${P2} = \frac{2*10^5*350}{300}$

P2 = 2.33 × $10^{5}$ Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × $10^{5}$ - 1.0 × $10^{5}$

gauge pressure = 1.33 × $10^{5}$ Pa

so gauge pressure is 133 kPa

4 0

3 years ago