A satellite at a distance of 36,000 km from an earth station radiates a power of 10 W from an

A construction manager is looking for workers to build a series of walkways within a large garden apartment. What type of workme

Lesechka [4]

Answer:

construction workers

Explanation:

A construction worker is usually someone with the technical skills and abilities needed to manually construct physical infrastructures.

Since a walkway is an infrastructure, hiring skilled construction workers should perform the project of building the series of walkways within the large garden apartment.

3 0

3 years ago

Which of the following is an example of a hardwood? A maple B spruce C pine D fir

bearhunter [10]

Answer:

A. Maple

Explanation:

Maple is a hardwood.

Hope that helps!

7 0

1 year ago

Consider a cubic workpiece of rigid perfect plastic material with side length lo. The cube is deformed plastically to the shape

Taya2010 [7]

Answer: ε₁+ε₂+ε₃ = 0

Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-

l₀l₀l₀=l₁l₂l₃

$\frac{lo*lo*lo}{l1*l2*l3}=1.0$

taking natural log on both sides

$ln(\frac{(lo*lo*lo)}{l1*l2*l3})=ln(1)$

Considering the logarithmic Laws of division and multiplication :

ln(AB) = ln(A)+ln(B)

ln(A/B) = ln(A)-ln(B)

$ln(\frac{(l1)}{lo})*ln(\frac{(l2)}{lo})*ln(\frac{(l3)}{lo}) = 0$

Use the image attached to see the definition of true strain defined as

ln(l1/1o)= ε₁

which then proves that ε₁+ε₂+ε₃ = 0

8 0

2 years ago

The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high sc

a_sh-v [17]

Answer:

(a)

dQ = mdq

dq = $C_p$ dT

$q = \int\limits^{T_2}_{T_1} {C_p} \, dT$ = $C_p$ (T₂ - T₁)

From the above equations, the underlying assumption is that $C_p$ remains constant with change in temperature.

(b)

Given;

V = 2L

T₁ = 300 K

Q₁ = 16.73 KJ , Q₂ = 6.14 KJ

ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter

Let $C_{cal}$ be heat constant of calorimeter

Q₂ = $C_{cal}$ ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m $C_p$ ΔT

number of moles of n-C₆H₁₄, n = m/M

ρ = 650 kg/m³ at 300 K

M = 86.178 g/mol

m = ρv = 650 (2x10⁻³) = 1.3 kg

n = m/M => 1.3 / 0.086178 = 15.085 moles

Q₁ - Q₂ = m $C_p$ ' ΔT

$C_p$ = (16.73 - 6.14) / (15.085 x 3.10)

$C_p$ = 0.22646 KJ mol⁻¹ k⁻¹

6 0

2 years ago

In a production facility, 3 cm thick large brass plates (k = 110 W/mC, α = 33.9 × 10-6 m2 /s) that are initially at a uniform

zysi [14]

Answer:

Explanation:

Given.

Thickness of brass plate $t = 3 cm$

Thermal conductivity of brass $k = 110 W/m.°C$

Density of brass $\rho = 8530 kg/m^3$

Specific heat of brass $C_p =380J/kg.°C$

Thermal diffusivity of brass $\alpha = 33.9\times 10^{-6} m^2/s$

Temperature of oven $T_{\infty} = 700°C$

The initial temperature $T_i= 25°C$

Plate remain in the oven $t =10 min$

Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness and there is thermal symmetry about the center plane.

The thermal properties of the plate are constant.

The heat transfer coefficient is constant and uniform over the entire surface.

The Fourier number is > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified).

The Biot number for this process $Bi = \frac{hL}{k}\\\\Bi=\frac{(80 W/m^2.°C)(0.015 m)}{(110 W/m.°C)}\\=Bi =0.0109$

The constants $\lambda_1$ and $A_1$ corresponding to this Biot are, from 11-2 tables.

The interpolation method used to find the

$\lambda_1=0.1039$ and $A_1=1.0018
$

The Fourier number $\tau=\frac{\alpha t}{L^2}\\\\\tau=\frac{(33.9\times 10^{-6} m^2/s)(10 min \times 60 s/min)}{(0.015m)^2}
\\\\\tau=90.4>0.2$

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable.

Then the temperature at the surface of the plates becomes

$\theta(L,t)_{wall}=\frac{T(x,t)-T_{\infty}}{(T_i-T_{\infty})}\\\\\theta(L,t)_{wall}=A_1e^{-\lambda_1^2\tau}\cos(\lambda_1L/L)\\\\\theta(L,t)_{wall}=(1.0018)e^{-(0.1039^2(90.4))}\cos(0.1039)\\\\\theta(L,t)_{wall}=0.378\\\\\frac{T(L,t)-700}{25-700}=0.378\\\\T(L,t)=445°C$

3 0

2 years ago

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