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Katena32 [7]
3 years ago
11

Steel bar A of 5 mm diameter is subject to a stress of 500 MPa. Aluminum bar B of 10 mm diameter is subject to a stress of 150 M

Pa. Is A or B bearing a higher load
Engineering
1 answer:
weqwewe [10]3 years ago
6 0

Answer:

aluminum bar carrying a higher load than steel bar

Explanation:

Given data;

steel abr

diameter = 5 mm

stress = 500 MPa

aluminium bar

diameter = 10 mm

stress = 150 MPa

we know

stress = laod/area

for steel bar

500 = \frac{P}{\frac{\pi}{4} 5^2}

solving for P

P = 9817.47 N

for Aluminium bar

150 = \frac{P}{\frac{\pi}{4} 10^2}

solving for P

P = 11790 N

aluminum bar carrying a higher load than steel bar

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Problem 3.10 One/half million parts of a certain type are to be manufactured annually on dedicated production machines that run
goldenfox [79]

Answer:

  • 4 machines
  • 4 machines

Explanation:

(a) The production schedule is 4000 hour per year, or 240,000 minutes. In that time, 133,333 1/3 parts can be produced. The production of 500,000 parts in a year will require the use of ...

  .5 million/(.1333... million/machine) = 3.75 machines

The demand can be satisfied by 4 machines.

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(b) If availability is 95% then 3.75/0.95 ≈ 3.95 machines will be required.

The demand can be satisfied by 4 machines.

8 0
3 years ago
Read 2 more answers
Homes may be heated by pumping hot water through radiators. What mass of water (in g) will provide the same amount of heat when
Nitella [24]

Answer:

a mass of water required is mw= 1273.26 gr = 1.27376 Kg

Explanation:

Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:

Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L

where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation

therefore

mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )

replacing values

mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg

3 0
4 years ago
Applications of fleming hand rule
Eva8 [605]

Answer:

Fleming hand rule represents the direction of current in a generator's windings and induced current as a conductor is attached to a circuit such that it moves in a magnetic field.

Explanation:

Fleming hand rule represents the direction of current in a generator's windings and induced current as a conductor is attached to a circuit such that it moves in a magnetic field.

Fleming hand rule is used in the case of electric motors and electric generators.

Fleming hand rule is used to determine the following:

1. Direction of torque

2. Angular velocity

3. Angular acceleration

4 0
3 years ago
If the rotational speed of a pump motor is reduced by 35%, what is the effect on the pump performance in terms of capacity, head
FinnZ [79.3K]

Answer:

- the capacity of the pump reduces by 35%.

- the head gets reduced by 57%.

the power consumption by the pump is reduced by 72%

Explanation:

the pump capacity is related to the speed as speed is reduces by 35%

so new speed is (100 - 35) = 65% of orginal speed

speed Q ∝ N ⇒ Q1/Q2 = N1/N2

Q2 = (N2/N1)Q1    

Q2 = (65/100)Q1

which means that the capacity of the pump is also reduces by 35%.

the head in a pump is related by

H ∝ N² ⇒ H1/H2 = N1²/N2²

H2 = (N2N1)²H1

H2 = (65/100)²H1 = 0.4225H1

so the head gets reduced by 1 - 0.4225 = 0.5775 which is 57%.

Now The power requirement of a pump is related as

P ∝ N³ ⇒ P1/P2 = N1³/N2³

P2 = (N2/N1)³P1

H2 = (65/100)²P1 = 0.274P1

So the reduction in power is 1 - 0.274 = 0.725 which is 72%

Therefore for a reduction of 35% of speed there is a reduction of 72% of the power consumption by the pump.  

8 0
3 years ago
Experimental Design Application Production engineers wish to find the optimal process for etching circuit boards quickly. They c
Veseljchak [2.6K]

Answer:

Hello your question is incomplete attached below is the missing part and answer

options :

Effect A

Effect B

Effect C

Effect D

Effect AB

Effect AC

Effect AD

Effect BC

Effect BD

Effect CD

Answer :

A  = significant

 B  = significant

C  = Non-significant

D  = Non-significant

AB  = Non-significant

AC  = significant

AD  = Non-significant

BC  = Non-significant

 BD  = Non-significant

 CD = Non-significant

Explanation:

The dependent variable here is Time

Effect of A  = significant

Effect of B  = significant

Effect of C  = Non-significant

Effect of D  = Non-significant

Effect of AB  = Non-significant

Effect of AC  = significant

Effect of AD  = Non-significant

Effect of BC  = Non-significant

Effect of BD  = Non-significant

Effect of CD = Non-significant

8 0
4 years ago
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