ENGINEERING COLLEGE

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is 74.96 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 105 GPa and 39.7 GPa, respectively.

Answers

Answer 1

Answer:

Answer:

L = 75.25 mm

Explanation:

First we need to find the lateral strain:

Lateral Strain = Change in Diameter/Original Diameter

Lateral Strain = (20.025 mm - 20 mm)/20 mm

Lateral Strain = 1.25 x 10⁻³

Now, we will find the Poisson's Ratio:

Poisson's Ratio = (E/2G) - 1

where,

E = Elastic Modulus = 105 GPa

G = Shear Modulus = 39.7 GPa

Therefore,

Poisson's Ratio = [(105 GPa)/(2)(39.7 GPa)] - 1

Poisson's Ratio = 0.322

Now, we find longitudinal strain by following formula:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

Longitudinal Strain = - Lateral Strain/Poisson's Ratio

Longitudinal Strain = - (1.25 x 10⁻³)/0.322

Longitudinal Strain = - 3.87 x 10⁻³

Now, we can fin the original length:

Longitudinal Strain = Change in Length/L

where,

L = Original Length = ?

Therefore,

- 3.87 x 10⁻³ = (74.96 mm - L)/L

(- 3.87 x 10⁻³)(L) + L = 74.96 mm

0.99612 L = 74.96 mm

L = 74.96 mm/0.99612

L = 75.25 mm

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. Were you able to observe ???? = 0 in the circuit you constructed during lab? Why or why not? Hint: What value of resistance would be needed for ???? = 0? 2. What feature in the time response of an RLC circuit distinguishes a critically damped response from an underdamped response? 3. Why must an op-amp be powered to be used in a circuit? 4. If you were handed a parts kit with an unknown op-amp, what information would you need to find prior to using it in a circuit?

Answers

Answer:

an attachment is below

Explanation:

1) the formula for damping coefficient id for RLC series circuit.

For \xi =0 you can make c=0 but inductor will still have some capacitance.

2) the responses of critically damped system and under damped system are shown with comments on their time response.

4) There can be many different answers to this question, but the 4 I have mentioned are the most important parameters we need to know about an unknown op-amp if we are to use it in our circuit.

Hope it answers all your questions.

A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust ends. The rocket has mass 2 kg and thrust force 35 N. Neglecting air resistance, determine (a) the speed of the rocket when the thrust ends, (b) the maximum height reached by the rocket, and (c) the speed of the rocket when it returns to the ground.

Answers

Answer:

a) $v=19.6 m/s$

b) $H=19.58 m$

c) $v_(f)=29.57 m/s$

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

$W=F_(tot)h=(F_(thrust)-mg)h=(35-(2*9.81))*25=384.5 J$

But we know the work is equal to change of kinetic energy, so:

$W=\Delta K=(1)/(2)mv^(2)$

$v=\sqrt{(2W)/(m)}=19.6 m/s$

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

$v_(f)^(2)=v_(i)^(2)-2gh$

At the maximum height the velocity is 0, so v(f) = 0.

$0=v_(i)^(2)-2gH$

$H=(19.6^(2))/(2*9.81)=19.58 m$

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

$v_(f)^(2)=v_(i)^(2)-2gh$

$v_(f)=\sqrt{19.6^(2)-(2*9.81*(-25))}=29.57 m/s$

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

A solid cylindrical workpiece made of 304 stainless steel is 150 mm in diameter and 100 mm is high. It is reduced in height by 50%, at room temperature, by open-die forging with flat dies. Assume that the coefficient of friction is 0.2. Calculate the forging force at the end of the stroke.

Answers

The answer is "45.3 NM".

There at end of the movement, the forging force is given by

$\to F = Y * \pi * r^2 * [1 + ((2 \mu r)/(3h))]$

h is the final height.

$\to h = (100)/(2)= 50 \ mm$

The ultimate radius is determined by following a volume constancy law, which states that volumes before deformation measured amount following distortion.

$\to \pi * 75^2 * 2 * 100 = \pi * r^2 * 2 * 50\n\n\to 75^2 * 2 = r^2\n\n\to r^2 = 11250\n\n\to r = √(11250)\n\n\to r = 106 \ mm\n\n\to E = \In((100)/(50))\n\n\to E = 0.69\n\n$

You may deduce from the graph flow that $Y = 1000\ MPa$ , thus we use the formula.

$= 1000 * 3.14 * 0.106^2 * [1 + (( 2 * 0.2 * 0.106)/(3 * 0.05))]\n\n= 1000 * 3.14 * 0.011236 * [1 + (( 0.0424)/(0.15))]\n\n= 35.3 * 1.2826\n\n = 45.3 \ MN\n\n\n$

Therefore, the answer is "45.3 NM".

Learn more:

brainly.com/question/17139328

Answer:

45.3 MN

Explanation:

The forging force at the end of the stroke is given by

F = Y.π.r².[1 + (2μr/3h)]

The final height, h is given as h = 100/2

h = 50 mm

Next, we find the final radius by applying the volume constancy law

volumes before deformation = volumes after deformation

π * 75² * 2 * 100 = π * r² * 2 * 50

75² * 2 = r²

r² = 11250

r = √11250

r = 106 mm

E = In(100/50)

E = 0.69

From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula

F = Y.π.r².[1 + (2μr/3h)]

F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]

F = 35.3 * [1 + 0.2826]

F = 35.3 * 1.2826

F = 45.3 MN

In a TDM communication example, 15 voice signals are badlimited to 5kHz and transmitted simultaneously using PAM. What is a preliminary estimate for the required system bandwidth?(a) 10 kHz
(b) 75 kHz
(c) 80 kHz
(d) 160 kHz
(e) None of the above.

Answers

Answer:

Option D

160 kHz

Explanation:

Since we must use at least one synchronization bit, total message signal is 15+1=16

The minimum sampling frequency, fs=2fm=2(5)=10 kHz

Bandwith, BW required is given by

BW=Nfs=16(10)=160 kHz

A test bottle containing just seeded dilution water has its DO level drop by 0.8 mg/L in a 5-day test. A 300 mL BOD bottle filled with 30 mL of wastewater and the rest with seeded dilution water experiences a drop of 7.3 mg/L in the same period (5-day). Calculate the BOD5 of the wastewater.

Answers

Answer:

BOD_5 = =65.8 mg/l

Explanation:

dilution water DO level = 0.8 m/l

BOD level drop to 7.3 mg/l

we know that BOD at 5th day can be clculated by using following relation $BOD_5 = ((D_1 -D_2) -(B_1-B_2)(1-P))/(P)$

$D_1 -D_2$ - DO drop in BOD bottle

$B_1-B_2$ - dilution water drop

P= 30/300 = 0.1

$BOD_5 = ((7.3) -(0.8)(1-0.1))/(0.1)$

$BOD_5 = =65.8 mg/l$

5. Switch a in the circuit has been open for a long time and switch b has been closed for a long time. Switch a is closed at t = 0. After remaining closed for 1s, switches a and b are opened simultaneously and remain open indefinitely. Determine the expression for the inductor current i that is valid when (a) 0 ≤ t ≤ 1s and (b) t ≥ 1s

Answers

Answer:

(a) 1/L∫Vdt; integral t [0,1]

(b) 1/L∫Vdt; integral t [ 1, infinity]

Explanation:

An Inductor current I, flowing through an inductor depends on the voltage, V, across the inductor and the inductance, L, of the inductor. The switch 1, 2 timing varies the voltage V with time t

The expression for inductor current is given as:

I= 1/L∫Vdt,

where I is equal to the current flowing through the inductor, L is equal to the inductance of the inductor, and V is equal to the voltage across the inductor.

The formula can also be written as:

I= I0 + 1/L∫Vdt, where I is inductor current at time t, and io is inductor current at t = 0. Time can be varied by controlling the switch

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