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3 years ago

Much of the workd went to bed hungry

1 answer:

7 0

The workers went to bed hungry probably because they are hard workers and so didn’t want to eat because they didn’t want to take break┌(;￣◇￣)┘

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. A line shaft rotating at 200 r.p.m. is to transmit 20 kW. The allowable shear stress for the material of the shaft is 42 MPa.

lbvjy [14]

Answer:

diameter=53.4 mm to 55 mm

Explanation:

7 0

3 years ago

1. Two aluminium strips and a steel strip are to be bonded together to form a composite bar. The modulus of elasticity of steel

yarga [219]

Answer:

1.933 KN-M

Explanation:

Determine the largest permissible bending moment when the composite bar is bent horizontally

Given data :

modulus of elasticity of steel = 200 GPa

modulus of elasticity of aluminum = 75 GPa

Allowable stress for steel = 220 MPa

Allowable stress for Aluminum = 100 MPa

a = 10 mm

First step

determine moment of resistance when steel reaches its max permissible stress

next : determine moment of resistance when Aluminum reaches its max permissible stress

Finally Largest permissible bending moment of the composite Bar = 1.933 KN-M

attached below is a detailed solution

5 0

3 years ago

Which professional draws maps and plans for projects involving structures other than buildings, such as bridges?

kolbaska11 [484]

Answer:

I believe it is a civil drafter

Explanation:

O*NET site

3 0

3 years ago

Read 2 more answers

What type of engineer would be most likely to develop a design for cars? chemical civil materials mechanical

Andreyy89

I don’t know but good luck

4 0

3 years ago

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

AURORKA [14]

Answer:

a) $w_{out} = 281.55\,\frac{kJ}{kg}$ , b) $s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}$

Explanation:

a) The process within the turbine is modelled after the First Law of Thermodynamics:

$-q_{out} - w_{out} + h_{in}-h_{out} = 0$

$w_{out} = h_{in} - h_{out}-q_{out}$

$w_{out} = c_{p}\cdot (T_{in}-T_{out})-q_{out}$

$w_{out} = \left(1.005\,\frac{kJ}{kg\cdot K}\right)\cdot (980\,K-670\,K)-30\,\frac{kJ}{kg}$

$w_{out} = 281.55\,\frac{kJ}{kg}$

b) The entropy production is determined after the Second Law of Thermodynamics:

$-\frac{q_{out}}{T_{surr}} + s_{in}-s_{out} + s_{gen} = 0$

$s_{gen} = \frac{q_{out}}{T_{surr}}+s_{out}-s_{in}$

$s_{gen} = \frac{q_{out}}{T_{surr}}+c_{p}\cdot \ln\left(\frac{T_{out}}{T_{in}} \right)$

$s_{gen} = \frac{30\,\frac{kJ}{kg} }{315\,K} + \left(1.005\,\frac{kJ}{kg\cdot K} \right)\cdot \ln\left(\frac{980\,K}{670\,K} \right)$

$s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}$

3 0

3 years ago