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3 years ago

The mean of 10 numbers is 9, then the sum (total) of these numbers will be

Engineering

2 answers:

Triss [41]3 years ago

6 0

Answer:

Explanation:

mean=sum÷number

sum=number×mean

sum=10×9

sum=90

qwelly [4]3 years ago

5 0

Answer:

Explanation:

mean is basically taking the sum of all numbers and then dividing the sum with the number of all given numbers..

here, the mean is 9, total numbers are 10.. so the sum will be 9 multiplied by 10, that is 90.

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Statement true about fats in food

Elodia [21]

Answer:

What that means please explain

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3 years ago

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A 30 mm thick AISI 1020 steel plate is sandwiched between two 10 mm thick 2024-T3 aluminum plates and compressed with a bolt and

denis-greek [22]

Answer:

275 MPa

Explanation:

Regardless of what it is holding, the stiffness of a bolt depends on its own material properties and geometry.

The stiffness is:

$k = E * \frac{A}{l}$

I assume this one is made of steel, because regular bolts are steel.

The Young's modulus for steel is E = 210 GPa

The longitude is given. (But note that in a real application you have to consider the length up to the nut.)

The section is (using the nominal diameter of 10 mm)

$A = \frac{\pi * d^2}{4} = \frac{\pi * 0.01^2}{4} = 7.85e-5 m^2$

Then:

$k = 2.1e11 * \frac{7.85e-5}{0.06} = 275e6 Pa = 275 MPa$

5 0

3 years ago

A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m ? K), and the wire/sheath interface

Svet_ta [14]

Question

A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m.K), and the wire/sheath interface is characterized by a thermal contact resistance of Rtc = 3E-4m².K/W. The convection heat transfer coefficient at the outer surface of the sheath is 10 W/m²K, and the temperature of the ambient air is 20°C.

If the temperature of the insulation may not exceed 50°C, what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?

Answer:

a. 4.52W/m

b. 13mm

Explanation:

Given

Diameter of electrical wire = 2mm

Wire Thickness = 2-mm

Thermal Conductivity of Rubberized sheath (k = 0.13 W/m.K)

Thermal contact resistance = 3E-4m².K/W

Convection heat transfer coefficient at the outer surface of the sheath = 10 W/m²K,

Temperature of the ambient air = 20°C.

Maximum Allowable Sheet Temperature = 50°C.

From the thermal circuit (See attachment), we my write

E'q = q' = (Tin,i - T∞)/(R'cond + R'conv)

= (Tin,i - T∞)/(Ln (r in,o / r in,i)/2πk + (1/(2πr in,o h)))

Where r in,i = D/2

= 2mm/2

= 1 mm

= 0.001m

r in,o = r in,i + t = 0.003m

T in, i = Tmax = 50°C

Hence

q' = (50 - 20)/[(Ln (0.003/0.001)/(2π * 0.13) + 1/(2π * 0.003 * 10)]

= 30/[(Ln3/0.26π) + 1/0.06π)]

= 30/[(1.34) + 5.30)]

= 30/6.64

= 4.52W/m

The critical radius is unaffected by the constant resistance.

Hence

Critical Radius = k/h

= 0.13/10

= 0.013m

= 13mm

5 0

3 years ago

water flows in a horizontal constant-area pipe; the pipe diameter is 75 mm and the average flow speed is 5 m/s. At the pipe inle

Veronika [31]

Answer:

Head loss = 28.03 m

Explanation:

According to Bernoulli's theorem for fluids we have

$\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=Constant$

Applying this between the 2 given points we have

$\frac{P_{1}}{\gamma _{w}}+\frac{V_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\gamma _{w}}+\frac{V_{2}^{2}}{2g}+z_{2}+h_{l}$

Here $h_{l}$ is the head loss that occurs

$\therefore h_{l}=\frac{P_{1}}{\gamma _{w}}+\frac{V_{1}^{2}}{2g}+z_{1}-\frac{P_{2}}{\gamma _{w}}-\frac{V_{2}^{2}}{2g}-z_{2}$

Since the pipe is horizantal we have $z_{1}-z_{2}=0$

Applying contunity equation between the 2 sections we get

$A_{1}V_{1}=A_{2}V_{2}\\\\\therefore V_{1}=V_{2}(\because A_{1}=A_{2})$

Since the cross sectional area of the both the sections is same thus the speed

is also same

Using this information in the above equation of head loss we obtain

$h_{l}=\frac{1}{\gamma _{w}}(P_{1}-P_{2})$

Applying values we get

$h_{l}=\frac{1}{9810}\times (275\times 10^{3})m\\\\\therefore h_{l}=28.03m$

3 0

3 years ago

A common procedure for measuring the velocity of an air stream involves insertion of an electrically heated wire (called a hot-w

timurjin [86]

Answer:

V = 6.33 m/s

Explanation:

Given:

- The length of the wire L = 0.02 m

- The diameter of the wire D = 0.0005 m

- The calibration expression V = 0.0000625*h^2

- Environment temperature T_inf = 298 K

- Surface temperature T_s = 348 K

- The voltage drop dV = 5 V

- The electric current I = 0.1 A

Find:

- the velocity of Air

Solution:

- Calculate the surface area of the wire:

A = pi*D*L

A = pi*(0.0005)*(0.02) = 0.00003142 m^2

- The rate of energy in the wire P:

P = I*dV = 0.1*5 = 0.5 W

- Apply Newton's Law of Cooling:

P = h*A*(T_s - T_inf)

h = P /A*(T_s - T_inf)

Plug in the values:

h= 0.5/ 0.00003142*(348 - 298)

h = 318.27 W /m^2K

- Using the calibration relationship given, compute the velocity of air:

V = 6.25*10^-5 * h^2

V = 6.25*10^-5 * (318.27)^2

V = 6.33 m/s

5 0

3 years ago

The mean of 10 numbers is 9, then the sum (total) of these numbers will be​

The mean of 10 numbers is 9, then the sum (total) of these numbers will be