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MArishka [77]
3 years ago
14

Question 3 (1 point)

Engineering
1 answer:
lord [1]3 years ago
7 0

Answer:

axonomeritric projection

this is the right answer

hope this is helpful

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You have a piece of 1045 steel 80 cm long and 30 cm wide and 15cm high. In this piece it is desired to make a groove along the 8
tino4ka555 [31]

Answer:

0.65 m/min

Explanation:

The volume of material to be removed is

80*8*10 = 650 cm^3

The tool has a diameter of 4 mm and a maximum axial cutting capacity of 50 mm, so its cross section normal to advance is

0.4*5 = 2 cm^2

If the groove have to be made in T = 5 minutes the advance speed would be

V/(S * T)

650/(2 * 5) = 65 cm/min = 0.65 m/min

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3 years ago
The _______ is a tendency to assume that people with one positive attribute (e.g., who are physically attractive) also have othe
morpeh [17]

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Halo effect

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3 years ago
A solid shaft is subjected to an axial load P = 200 kN and a torque T = 1.5 kN.m. a) Determine the diameter of the shaft if the
Rom4ik [11]

Answer:

<em>a) 42 mm</em>

<em>b) 144.4 MPa</em>

<em></em>

Explanation:

Load P = 200 kN = 200 x 10^3 N

Torque T = 1.5 kN-m = 1.5 x 10^3 N-m

maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa

diameter of shaft d = ?

From T = τ * \frac{\pi }{16} * d^{3}

substituting values, we have

1.5 x 10^3 = 100 x 10^6 x \frac{3.142 }{16} x d^{3}

d^{3} = 7.638 x 10^-5

d = \sqrt[3]{7.638 * 10^-5} = 0.042 m = <em>42 mm</em>

b) Normal stress = P/A

where A is the area

A = \frac{\pi d^{2} }{4} = \frac{3.142*0.042^{2} }{4} = 1.385 x 10^-3

Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = <em>144.4 MPa</em>

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4 years ago
To combat overheating, keep a ___________ in your car.
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Keep a gallon of water in your car.

Explanation:

Because.

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3 years ago
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Who is the first presidant of the us?
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George Washington was the first president

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