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3 years ago

Ten dollars per hour is about how much income per year

Engineering

2 answers:

Dafna1 [17]3 years ago

6 0

Depends how many hours you wanna work

IgorLugansk [536]3 years ago

5 0

Answer:

see below

Explanation:

10 per hour

example: 37.5 hours in a week

10 x 37.5 hours per week x 52 weeks in a year

19,500 per year.

you can use the above calculation depending on how much you want to work per week.

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If the specific surface energy for aluminum oxide is 0.90 J/m2 and its modulus of elasticity is (393 GPa), compute the critical

vampirchik [111]

Answer:

critical stress required for the propagation is 27.396615 × $10^{6}$ N/m²

Explanation:

given data

specific surface energy = 0.90 J/m²

modulus of elasticity E = 393 GPa = 393 × $10^{9}$ N/m²

internal crack length = 0.6 mm

to find out

critical stress required for the propagation

solution

we will apply here critical stress formula for propagation of internal crack

( σc ) = $\sqrt{\frac{2E\gamma s}{\pi a}}$ .....................1

here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 × $10^{-3}$ m

so now put value in equation 1 we get

( σc ) = $\sqrt{\frac{2E\gamma s}{\pi a}}$

( σc ) = $\sqrt{\frac{2*393*10^9*0.90}{\pi 0.3*10^{-3}}}$

( σc ) = 27.396615 × $10^{6}$ N/m²

so critical stress required for the propagation is 27.396615 × $10^{6}$ N/m²

6 0

3 years ago

REVIEW QUESTIONS

babunello [35]

Snap rings, and bearings can be used to keep a gear on a shaft, hope this helps!!

5 0

3 years ago

How to make text take shape of object in affinity designer

Alina [70]

Answer:

To fit text to a shape in Affinity Designer, make sure you have your text selected. Then, grab the Frame Text Tool and click on the shape. A blinking cursor will appear within the shape, indicating that you can begin typing. The text you type will be confined to the boundaries of the shape.

Explanation:

6 0

3 years ago

At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho

victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

$\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2} }{2})^{2}+\tau _{xy}^{2} }$

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

$32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2} }$

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

$\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}$

Where

σp1 = 20 ksi

σp2 = -30 ksi

$\sigma _{p1} +\sigma _{p2}=-10 ksi$ (equation 1)

$\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi$ equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0

4 years ago

A train consists of a 50 Mg engine and three cars, each having a mass of 30 Mg . If it takes 75 s for the train to increase its

ohaa [14]

Answer:

T = 15 kN

F = 23.33 kN

Explanation:

Given the data in the question,

We apply the impulse momentum principle on the total system,

mv₁ + ∑ $\int\limits^{t2}_{t1} {Fx} \, dt$ = mv₂

we substitute

[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )

F( 75 - 0 ) = 1.75 × 10⁶

The resultant frictional tractive force F is will then be;

F = 1.75 × 10⁶ / 75

F = 23333.33 N

F = 23.33 kN

Applying the impulse momentum principle on the three cars;

mv₁ + ∑ $\int\limits^{t2}_{t1} {Fx} \, dt$ = mv₂

[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )

F(75-0) = 1.125 × 10⁶

The force T developed is then;

T = 1.125 × 10⁶ / 75

T = 15000 N

T = 15 kN

7 0

3 years ago