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Jet001 [13]
3 years ago
14

A specimen of commercially pure copper has a strength of 240 MPa. Estimate its average grain diameter using the Hall-Petch equat

ion.
Engineering
1 answer:
romanna [79]3 years ago
5 0

Answer:

3.115× 10^{-3} meter

Explanation:

hall-petch constant for copper is given by

      S_0=25 MPa

      k=0.12 for copper

now according to hall-petch equation

S_Y=S_0 +\frac{K}{\sqrt{D}}

240=25+\frac{0.12}{\sqrt{D}}

D=3.115× 10^{-3} meter

so the grain diameter using the hall-petch equation=3.115×  10^{-3} meter

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A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 3600 rpm. When the power is turn
Ostrovityanka [42]

Answer: a) 150 rev. b) 2105 rev.

Explanation:

a) Assuming a uniformly accelerated motion, we can use the equivalent kinematic equations, replacing linear variables by angular ones.

In order to get the number of revolutions executed, we can use this:

ωf² - ω₀² = 2 γ Δθ (1)

For the first part, we know that ω₀ = 0 (as it starts from rest).

We can find out the value of angular acceleration γ, just applying the definition of angular acceleration, as the change in angular velocity, regarding time, as follows:

γ = (ωf - ω₀) / Δt (2)

As we would want to use SI units, it is advisable to convert the value of ωf, from rpm to rad/sec.

3600 rev/min . (1min/60 sec) . (2π rad/rev) = 120π rad/sec

Replacing in (2), we get γ:

γ = 120 π / 5 rad/sec² = 24 π rad/sec²

Replacing in (1) and solving for Δθ:

Δθ = 120² π² / 2. 24 π = 300 π rad

As 1 rev = 2π rad, Δθ = 150 rev

b) For the second part, we can use exactly the same equations, taking into account that ω₀ = 120 π rad/sec, and that ωf = 0.

The new value for γ is as follows:

γ = -120π  / 70 rad/sec² = -1.71 rad/sec²

Replacing in (1) and solving for Δθ, we get:

Δθ = -120² π² / 2. (-1.71) π = 4210 π rad

As 1 rev = 2π rad, Δθ = 2105 rev

7 0
4 years ago
Show that y = '(t - s)f(s)ds is a solution to my" + ky = f(t). Use g' (0) = 1/m and mg" + kg = 0. 6.1) Derive y' 6.2) Using g(0)
Gre4nikov [31]

Answer:

Explanation:

Given that:

y = \int^t_og'(t-s) f(s) ds \  \text{is  solution to } \ my"ky= f(t)

where;

g'(0) = \dfrac{1}{m}     and mg"+kg = 0

\text{Using Leibniz Formula to prove the above equation:}

\dfrac{d}{dt} \int ^{b(t)}_{a(t)} \ f (t,s) \ ds = f(t,b(t) ) * \dfrac{d}{dt}b(t) - f(t,a(t)) *\dfrac{d}{dt}a(t) + \int ^{b(t)}_{a(t)}\dfrac{\partial}{\partial t} f(t,s) \ dt

So, y = \int ^t_0  g' (t-s) f(s) \ ds

\text{By differentiation with respect to t;}

y' = g'(o) f(t) \dfrac{d}{dt}t- 0 + \int^{t}_{0}g'' (t-s) f(s) ds \\ \\  y' = \dfrac{1}{m}f(t) + \int ^t_0 g'' (ts) f(s) \ ds

y'' = \dfrac{1}{m} f'(t) + g"(0) f(t) + \int^t_o g"'(t-s) f(s)ds --- (1)

Since \ \ mg" (t) +kg (t) = 0  \\ \\  \implies g" (t) = -\dfrac{k}{m} g(t) --- (111) \\ \\  put \  t \  =0 \  we  \ get;\\g" (0) = - \dfrac{k}{m } g(0)  \\ \\  g"(0) = 0 \ \ \ \   ( because \  g(0) =0) \\ \\

Now \ differentiating \ equation (111) \ with \ respect \ to \ t  \\ \\  g"'(t) = -\dfrac{k}{m}g(t)  \\ \\  replacing  \ it \ into  \ equation \ (1) \\ \\ y" = \dfrac{1}{m}f' (t) + 0 + \int ^t_o  \dfrac{-k}{m}g' (t-s) f(s) \ ds \\ \\ y" = \dfrac{1}{m}f' (t) - \dfrac{k}{m} \int ^t_o g' (t-s) \ f(s) \ ds \\ \\  y" = \dfrac{1}{m}f'(t) - \dfrac{k}{m}y \\ \\  my" = f'(t)-ky \\ \\ \implies \mathbf{ my" +ky = f'(t)}

7 0
3 years ago
Consider the control volume form of the basic laws. For the conservation of mass form for a control volume with mass flow into a
aleksandrvk [35]

Answer:

C) Dependent on the mass flows in and out.

Explanation:

Lets take control volume(CV)

Take

          m_i =inlet mass flow rate

          m_e =exit mass flow rate

If we take unsteady flow process then inlet mass can not be equal to exit mass.Some mass can store if inlet mass flow rate is high and exit mass flow rate  is low.

So mass of control volume

    m_{cv}=m_i-m_e.

so above we can say that mass of control volume dependent on inlet and exit mass.

5 0
3 years ago
The pressure of a car can be adjusted using what​
IgorLugansk [536]

Answer:

tyre pressure monitoring system (TPMS)

Explanation:

It very important to inflate your tyres to the specified pressure. So cultivate a habit to check and refill them once a month. Note that you can't tell if a tyre is underinflated just by looking at it. If it actually looks underinflated in a way. And you can't really rely on a tyre pressure monitoring system (TPMS) only to keep track. Most systems only warn you when the pressure is 25 percent below the manufacturer's recommended tyre pressure.

3 0
3 years ago
What is the relationship between green building and self-interest? (Select all that apply.)
Lady bird [3.3K]

The relationship between green building and self-interest include the following:

  • Green building is burdensome but has advantages in terms of public image
  • Green building is attractive because of its long-term growth potential as a market.

<h3>What is Green building?</h3>

This is the process of reducing the environmental impact of buildings. This therefore leads to enhancement of the health of people.

Its relationship with self interest can seen listed above as the most appropriate choice.

Read more about Green building here brainly.com/question/9779427

#SPJ1

4 0
3 years ago
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