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2 years ago

Hello , how are yall:))))

Engineering

2 answers:

SVEN [57.7K]2 years ago

6 0

Answer:

eh I'm good hbu?????????

kumpel [21]2 years ago

4 0

Answer:

Tired,depressed, and feeling like a worm on a string

Explanation:

None

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What is shown in the above figure

Vikki [24]

Answer:

there is no figure .....

3 0

2 years ago

Most technician jobs in the field of metrology require a college degree. True or False?

Elanso [62]

Answer:

True

Explanation:

Because I Said

7 0

3 years ago

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simply supported beam is subjected to a linearly varying distributed load ( ) 0 q x x L 5 q with maximum intensity 0 q at B. The

Pavlova-9 [17]

Answer:

q₀ = 350,740.2885 N/m

Explanation:

Given

$q(x)=\frac{x}{L} q_{0}$

σ = 120 MPa = 120*10⁶ Pa

$L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\$

We can see the pic shown in order to understand the question.

We apply

∑MB = 0 (Counterclockwise is the positive rotation direction)

⇒ - Av*L + (q₀*L/2)*(L/3) = 0

⇒ Av = q₀*L/6 (↑)

Then, we apply

$v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x$

Then, we can get the maximum bending moment as follows

$M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\ x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m$

then we get

$M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}$

We get the inertia as follows

$I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}$

We use the formula

σ = M*y/I

⇒ M = σ*I/y

where

$y=\frac{h}{2} =\frac{0.3m}{2}=0.15m$

If M = Mmax, we have

$(\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4} }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}$

8 0

3 years ago

#5 Air undergoes an adiabatic compression in a piston-cylinder assembly from P1= 1 atm and Ti=70 oF to P2= 5 atm. Employing idea

otez555 [7]

Answer:

The work transfer per unit mass is approximately 149.89 kJ

The heat transfer for an adiabatic process = 0

Explanation:

The given information are;

P₁ = 1 atm

T₁ = 70°F = 294.2611 F

P₂ = 5 atm

γ = 1.5

Therefore, we have for adiabatic system under compression

$T_{2} = T_{1}\cdot \left (\dfrac{P_{2}}{P_{1}} \right )^{\dfrac{\gamma -1}{\gamma }}$

Therefore, we have;

$T_{2} = 294.2611 \times \left (\dfrac{5}{1} \right )^{\dfrac{1.5 -1}{1.5 }} \approx 503.179 \ K$

The p·dV work is given as follows;

$p \cdot dV = m \cdot c_v \cdot (T_2 - T_1)$

Therefore, we have;

Taking air as a diatomic gas, we have;

$C_v = \dfrac{5\times R}{2} = \dfrac{5\times 8.314}{2} = 20.785 \ J/(mol \cdot K)$

The molar mass of air = 28.97 g/mol

Therefore, we have

$c_v = \dfrac{C_v}{Molar \ mass} = \dfrac{20.785}{28.97} \approx 0.7175 \ kJ/(kg \cdot K)$

The work done per unit mass of gas is therefore;

$p \cdot dV =W = 1 \times 0.7175 \times (503.179 - 294.2611) \approx 149.89 \ kJ$

The work transfer per unit mass ≈ 149.89 kJ

The heat transfer for an adiabatic process = 0.

8 0

2 years ago

Explain the purpose of wrench plzzz

Rus_ich [418]

Answer:

Wrenches are made in various shapes and sizes and are used for gripping, fastening, turning, tightening and loosening things like pipes, pipe fittings, nuts and bolts. There are basically two major kinds of wrenches: Pipe wrenches used in plumbing for gripping round (cylindrical) things.

4 0

2 years ago

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