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3 years ago

What information in drawing's title block identifies the project?

Engineering

1 answer:

mezya [45]3 years ago

7 0

Answer:

B would b your answer

Explanation:

have a great day. please mark as brainllest

[o]-[o]

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Solve the compound inequality. 3x − 4 > 5 or 1 − 2x ≥ 7

Kaylis [27]

X ≤ −3 or x > 3
inequality form

3 0

3 years ago

A freshly annealed glass containing flaws of maximum length of 0.1 microns breaks under a tensile stress of 120 MPa. If a sample

almond37 [142]

Answer:

0.16 micron per day

Explanation:

Given:

The initial crack length, a₁ = 0.1 micron = 0.1 × 10⁻⁶ m

Initial tensile stress, σ₁ = 120 MPa

Final stress = 30 MPa

now from Griffith's equation, we have

$\sigma=[\frac{G_cE}{\pi\ a}]^\frac{1}{2}$

where,

Gc and E are the material constants

now,

for the initial stage

$120=[\frac{G_cE}{\pi\ (0.1\times10^{-6}}]^\frac{1}{2}$ ........{1}

and for the final case

$30=[\frac{G_cE}{\pi\ a_2}]^\frac{1}{2}$ ............{2}

on dividing 1 by 2, we get

$\frac{120}{30}=[\frac{a_2}{0.1\times10^{-6}}]^\frac{1}{2}$

a₂ = 4² × 0.1 × 10⁻⁶ m

a₂ = 1.6 micron

Now,

the change from 0.1 micron to 1.6 micron took place in 10 days

therefore, the rate at which the crack is growing = $\frac{1.6-0.1}{10}$

average rate of change of crack = 0.16 micron per day

6 0

3 years ago

What the phat is this

Alex17521 [72]

Answer:

It's Brainly ;)

8 0

3 years ago

Read 2 more answers

A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to c

Rus_ich [418]

Answer:

Heat required =7126.58 Btu.

Explanation:

Given that

Mass m=20 lb

We know that

1 lb =0.45 kg

So 20 lb=9 kg

m=9 kg

Ice at -15° F and we have to covert it at 200° F.

First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.

We know that

Specific heat for ice $C_p=2.03\ KJ/kg.K$

Latent heat for ice H=336 KJ/kg

Specific heat for ice $C_p=4.187\ KJ/kg.K$

We know that sensible heat given as

$Q=mC_p\Delta T$

Heat for -15F to 32 F:

$Q=mC_p\Delta T$

$Q=9\times 2.03(32+15) KJ$

Q=858.69 KJ

Heat for 32 Fto 200 F:

$Q=mC_p\Delta T$

$Q=9\times 4.187(200-32) KJ$

Q=6330.74 KJ

Total heat=858.69 + 336 +6330.74 KJ

Total heat=7525.43 KJ

We know that 1 KJ=0.947 Btu

So 7525.43 KJ=7126.58 Btu

So heat required to covert ice into water is 7126.58 Btu.

8 0

3 years ago

A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.

Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = $\frac{PL}{AE}$ ................1

here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = $\frac{PL}{AE}$

δ = $\frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}$

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0

3 years ago