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3 years ago

1. Drill press size is determined by the largest__

Engineering

1 answer:

hodyreva [135]3 years ago

3 0

Drill press size is determined by the largest piece of stock

Your welcome

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3 years ago

Water is boiled in a pot covered with a loosely fitting lid at a location where the pressure is 85.4 kPa. A 2.61 kW resistance h

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Answer:

t = 6179.1 s = 102.9 min = 1.7 h

Explanation:

The energy provided by the resistance heater must be equal to the energy required to boil the water:

E = ΔQ

ηPt = mH

where.

η = efficiency = 84.5 % = 0.845

P = Power = 2.61 KW = 2610 W

t = time = ?

m = mass of water = 6.03 kg

H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Therefore,

(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)

$t = \frac{1.362\ x\ 10^7\ J}{2205.45\ W}$

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2 years ago

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3 years ago

B/ Evaluate e^(πi/2)

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2 years ago

Read 2 more answers

• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the

Rudik [331]

Answer:

minimum factor of safety for fatigue is = 1.5432

Explanation:

given data

AISI 1018 steel cold drawn as table

ultimate strength Sut = 63.800 kpsi

yield strength Syt = 53.700 kpsi

modulus of elasticity E = 29.700 kpsi

we get here

$\sigma a$ = $\sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}$ ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

$\sigma a$ = $\sqrt{(12 \times 1)^2+3\times (0\times 1)^2}$

$\sigma a$ = 12 kpsi

and

$\sigma m$ = $\sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}$ ...........2

put here value and we get

$\sigma m$ = $\sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}$

$\sigma m$ = 17.34 kpsi

now we apply here goodman line equation here that is

$\frac{\sigma m}{Sut} + \frac{\sigma a}{Se} = \frac{1}{FOS}$ ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

$\frac{17.34}{63.800} + \frac{12}{31.9} = \frac{1}{FOS}$

solve it we get

FOS = 1.5432

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3 years ago