The title block generally contains _______

What’s the most important benefit of maintaining a neutral posture

olga55 [171]

Neutral posture is essential for optimal wellbeing and functioning of the body. Holding the weight of the body The most important function of a neutral posture is to maintain the body in an upright position, supporting the body against gravity

7 0

3 years ago

I’m bored let’s text

Nuetrik [128]

Answer:

hi same here. hru btw

안녕하세요 모두가 잘되기를 바랍니다. 안전 유지

3 0

3 years ago

Read 2 more answers

Calculate KI for a rectangular bar containing an edge crack loaded in three-point bending where P=35.0 kN, W=50.8 mm, B=25 mm, a

Katyanochek1 [597]

Answer:

$K_{I}=5.21 MPa\sqrt{m}$

Explanation:

given data

Load P = 35 kN

Width of bar W = 50.8 mm

Breadth of bar B = 25 mm

Ratio of crack length to width α = a/W = 0.2

solution

we get here KI for a rectangular bar that is express as

$K_{I} = \frac{6P}{BW}Y\sqrt{\pi a}$ ................................1

here Y is the geometrical function

so

Y = $\frac{1.12+\alpha (3.43\alpha -1.89)}{1-0.55\alpha}$

Y = $\frac{1.12+0.2(3.43\times 0.2-1.89)}{1-0.55\times 0.2}$

Y = $\frac{0.8792}{0.89}$

Y = 0.9878

so put here value in equation 1

$K_{I} = \frac{6\times 35\times 10^{3}}{0.025\times 0.0508}\times 0.9878\times \sqrt{3.1415\times (0.2\times 0.0508)}$

$K_{I} = 165354.33\times 10^{3}\times 0.9878\times 0.0319$

$K_{I}$ = 5210.45 × 10³ $Pa\sqrt{m}$

$K_{I}$ = 5.21 MPa $\sqrt{m}$

5 0

3 years ago

Heat is transferred at a rate of 2 kW from a hot reservoir at 875 K to a cold reservoir at 300 K. Calculate the rate at which th

blsea [12.9K]

Answer:

The correct answer is 0.004382 kW/K, the second law is a=satisfied and established because it is a positive value.

Explanation:

Solution

From the question given we recall that,

The transferred heat rate is = 2kW

A reservoir cold at = 300K

The next step is to find the rate at which the entropy of the two reservoirs changes is kW/K

Given that:

Δs = Q/T This is the entropy formula,

Thus

Δs₁ = 2/ 300 = 0.006667 kW/K

Δs₂ = 2 / 875 =0.002285

Therefore,

Δs = 0.006667 - 0.002285

= 0.004382 kW/K

Yes, the second law is satisfied, because it is seen as positive.

8 0

4 years ago

A 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves throu

Drupady [299]

Answer:

The temperature of the strip as it exits the furnace is 819.15 °C

Explanation:

The characteristic length of the strip is given by;

$L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m$

The Biot number is given as;

$B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952$

$B_i$ < 0.1, thus apply lumped system approximation to determine the constant time for the process;

$\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s$

The time for the heating process is given as;

$t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s$

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

$T(t) = T_{ \infty} + (T_i -T_{\infty})e^{-t/ \tau}\\\\T(t) = 930 + (20 -930)e^{-300/ 142.5}\\\\T(t) = 930 + (-110.85)\\\\T_{(t)} = 819.15 \ ^0 C$

Therefore, the temperature of the strip as it exits the furnace is 819.15 °C

5 0

3 years ago

The title block generally contains ________.