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3 years ago

9

A steel plate has a hole drilled through it. The plate is put into a furnace and heated. What happens to the size of the inside

diameter of a hole as its temperature increases

1 answer:

djverab [1.8K]3 years ago

4 0

Answer:

The diameter increases

Explanation:

The expansion in the metal is uniform in every dimension

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Answer:

solar engineering field

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3 years ago

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Q7. A cylindrical rod of 1040 steel originally 15.2 mm (0.60 in.) in diameter is to be cold worked by drawing; the circular cros

umka2103 [35]

Answer:

11.2mm or 0.45in

Explanation:

The percent cold work, attendant tensile strength and ductility if drawing is carried out without interruption is given by the equation you will find in the attached file.

Please go through the attached file for a step by step solution to this question.

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3 years ago

Consider the velocity boundary layer profile for flow over u flat plate to be of the form u = C_1 + C_2 y. Applying appropriate

ra1l [238]

Answer:

The result in terms of the local Reynolds number ⇒ Re = [μ_∞ · x] / v

Explanation:

See below my full workings so you can compare the results with those obtained from the exact solution.

4 0

4 years ago

The drag coefficient of a car at the design conditions of 1 atm, 25°C, and 90 km/h is to be determined experimentally in a large

SIZIF [17.4K]

Answer: 0.288

Explanation:

Given

Pressure of the car, P = 1 atm

Temperature of the car, T = 25° C

Speed of the car, v = 90 km/h = 90*1000/3600 = 25 m/s

Height of the car, h = 1.25 m

Width of the car, b = 1.65 m

Force acting on the far, F = 220 N

Drag coefficient, C(d) = ?

Using our table A-9, we can trace that the density of air ρ, at the given temperature and pressure of 25 °C and 1 atm, is 1.184 kg/m³

Area = h *b

Area = 1.25 * 1.65

Area = 2.0625 m²

Now we solve for the drag coefficient using the formula

C(d) = F / (1/2 * ρ * A * v²)

C(d) = 220 / (0.5 * 1.184 * 2.0625 * 25²)

C(d) = 220 / (1.221 * 625)

C(d) = 220 / 763.125

C(d) = 0.288

Therefore, the drag coefficient is 0.288

3 0

3 years ago

he mean weight of a breed of yearling cattle is 11871187 pounds. Suppose that weights of all such animals can be described by th

Tamiku [17]

Question

The mean weight of a breed of yearling cattle is 1187 pounds. Suppose that weights of all such animals can be described by the Normal model N(1187,78).

a) How many standard deviations from the mean would a steer weighing 1000 pounds be?

b) Which would be more unusual, a steer weighing 1000 pounds, or one weighing 1250 pounds?

Answer:

a. z = -2.40

A sleet weighing 1,000 pounds is 2.40 standard deviations below the mean.

b. z = 0.81

1000 is more unusual because its contained on the extreme end from the mean

Explanation:

a.

Let weight (in pounds) of the cattle be denoted by letter x:

z = (x - u)/ σ

Where u = mean and σ = standard deviation

u = 1187

σ = 78

x = 1000

Use z score formula to standardize the value of x:

z = (1000 - 1187)/78

z = -187/78

z = -2.397436

z = -2.40 ------_ Approximated

A sleet weighing 1,000 pounds is 2.40 standard deviations below the mean.

b.

x= 1250

z= (1250 - 1187)/78

z = 63/78

z = 0.807692

z = 0.81 --------- Approximated

1000 is more unusual because its contained on the extreme end from the mean

8 0

3 years ago