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vitfil [10]
2 years ago
5

Shops should avoid purchasing any material sold in ____________.

Engineering
1 answer:
snow_lady [41]2 years ago
7 0

Answer: Aerosol Cans

Explanation: I just did the quiz

You might be interested in
A 10-m long steel linkage is to be designed so that it can transmit 2 kN of force without stretching more than 5 mm nor having a
Fed [463]

Answer:

<em>minimum required diameter of the steel linkage is 3.57 mm</em>

<em></em>

Explanation:

original length of linkage l = 10 m

force to be transmitted  f = 2 kN = 2000 N

extension e = 5 mm= 0.005 m

maximum stress σ = 200 N/mm^2 = 2*10^{8}  N/m^{2}

maximum stress allowed on material σ = force/area

imputing values,

200 = 2000/area

area = 2000/(2*10^{8}) = 10^{-5} m^2

recall that area = \pi d^{2} /4

10^{-5} = \frac{3.142*d^{2} }{4} = 0.7855d^{2}

d^{2} = \frac{10^{-5} }{0.7855} = 1.273*10^{-5}

d = \sqrt{1.273*10^{-5}  } = 3.57*10^{-3} m = 3.57 mm

<em>maximum diameter of  the steel linkage d = 3.57 mm</em>

4 0
2 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in
RideAnS [48]

To resolve this problem we have,

R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm

F_{f2} is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa

After that, we can calculate the flexural strenght for the square cross section using the previously value.

\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN

6 0
2 years ago
A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses
yaroslaw [1]

Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature T_1 = 25° C

T{\infty} =600^0C

h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}

Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}

Bi = \dfrac{2.5}{231}

Bi = 0.0108

The time constant value \tau_t is :

\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}

\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}

\tau_t = \dfrac{2702* 0.025*1033}{100}

\tau_t = 697.79

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]

where;

Q = -\Delta E _{st} which correlates with the change in the internal energy of the solid.

So;

Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}

The maximum value for the change in the internal energy of the solid  is :

(pVc)\theta_1 = -\Delta E _{st}max

By equating the two previous equation together ; we have:

\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{  (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

0.75=  [1-e^{\dfrac {-t}{ \tau_1}}]}

So;

0.75=  [1-e^{\dfrac {-t}{ 697.79}}]}

1-0.75=  [e^{\dfrac {-t}{ 697.79}}]}

0.25 =  e^{\dfrac {-t}{ 697.79}}

In(0.25) =  {\dfrac {-t}{ 697.79}}

-1.386294361= \dfrac{-t}{697.79}

t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}

\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}

\dfrac{T - 600}{25-600}= 0.25

\dfrac{T - 600}{-575}= 0.25

T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

3 0
2 years ago
What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.
yanalaym [24]

Answer:

Re=\dfrac{\rho\ v\ l}{\mu }

Explanation:

Reynolds number:

  Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is  low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

Re=\dfrac{F_i}{F_v}

For plate can be given as

Re=\dfrac{\rho\ v\ l}{\mu }

Where  ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.

Reynolds\ number\is \

7 0
2 years ago
Holmes owns two suits: one black and one tweed. He always wears either a tweed suit or sandals. Whenever he wears his tweed suit
vazorg [7]

Answer:

He wore his black suit, another color of shirt (not purple) and shoes

Explanation:

Holmes owns two suits: one black and one tweed.

Whenever he wears his tweed suit and a purple shirt, he chooses not to wear a tie and whenever he wears sandals, he always wears a purple shirt.

So, if he wore a bow tie yesterday, it means he wore his black suit, another color of shirt (not purple) and shoes because the shirt color is not purple

4 0
2 years ago
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