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olasank [31]
3 years ago
6

A flywheel accelerates for 5 seconds at 2 rad/s2 from a speed of 20 rpm. Determine the total number of revolutions of the flywhe

el during the period of its acceleration. a.5.65 b.8.43 c. 723 d.6.86
Engineering
1 answer:
s344n2d4d5 [400]3 years ago
7 0

Answer:

option (a)

Explanation:

t = 5 sec, α = 2 rad/s^2, f0 = 20 rpm = 20 / 60 rps

Use second equation of motion for rotational motion

θ = ω0 x t + 1/2 α t^2

θ = 2 x 3.14 x 5 x 20 / 60 + 0.5 x 2 x 5 x 5

θ = 10.47 + 25 = 35.47 rad

Number of revolution = 35.47 / (2 x 3.14) = 5.65

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The unit for volume flow rate is gallons per minute, but cubic feet per second is preferred. Use the conversion factor tables in
amm1812

Answer:

The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)

Explanation:

Since the given volume flow rate is gallons per minute.

We know that 1 gallon = 3.785 liters and

1 minute = 60 seconds

Let the flow rate be Q\frac{gallons}{minute}

Now replacing the gallon and the minute by the above values we get

Q'=Q\frac{gallon}{minute}\times \frac{3.785liters}{gallon}\times \frac{1minute}{60seconds}

Thus Q'=0.631Q\frac{liters}{second}

Now since we know that 1 liter = 0.0353ft^{3}

Using this in above relation we get

Q'=0.631Q\frac{liters}{second}\times \frac{0.0353ft^3}{liters}\\\\\therefore Q'=0.00223Q

From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.

3 0
3 years ago
Ti-6Al-4V has a fracture toughness of 74.6 MPa-m0.5. How much stress (in MPa) would it take to fail a plate loaded in tension th
Nikitich [7]

Answer:

critical stress  = 595 MPa

Explanation:

given data

fracture toughness =  74.6 MPa-\sqrt{m}

crack length = 10 mm

f = 1

solution

we know crack length = 10 mm  

and crack length = 2a as given in figure attach

so 2a = 10

a = 5 mm

and now we get here with the help of plane strain condition , critical stress is express as

critical stress  = \frac{k}{f\sqrt{\pi a}}    ......................1

put here value and we get

critical stress  = \frac{74.6}{1\sqrt{\pi 5\times 10^{-3}}}

critical stress  = 595 MPa

so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.

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6 0
3 years ago
Which of the following has led to a safer and more prosperous global community within the last century? the Bronze Age composite
sattari [20]

Answer:

The bronze age

4 0
3 years ago
Read 2 more answers
Which of the following is NOT true about hydraulic systems?
Dmitry [639]

Answer: The answer is D

D.In hydraulic systems, the operating temperatures must be kept between 170�F and 180°F 

Explanation:

The operating temperature for hydraulic systems is 140°F and below. Anything above this temperature is too high and will reduce the useful life of hydraulic fluid.

Most often problems associated with hydraulic systems are caused by fluid contaminated with particulate matter.

7 0
3 years ago
A utility generates electricity with a 36% efficient coal-fired power plant emitting the legal limit of 0.6 lb of SO2 per millio
Mama L [17]

Answer:

a) 570 kWh of electricity will be saved

b) the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c) $1.296 can be earned by selling the SO₂ saved by a single CFL

Explanation:

Given the data in the question;

a) How many kilowatt-hours of electricity would be saved?

first, we determine the total power consumption by the incandescent lamp

P_{incandescent} = 75 w × 10,000-hr = 750000 wh = 750 kWh

next, we also find  the total power consumption by the fluorescent lamp

P_{fluorescent} = 18 × 10000 = 180000 = 180 kWh

So the value of power saved will be;

P_{saved} = P_{incandescent}  - P_{fluorescent}

P_{saved} = 750 - 180

P_{saved}  = 570 kWh

Therefore, 570 kWh of electricity will be saved.

now lets find the heat of electricity saved in Bituminous

heat saved = energy saved per CLF / efficiency of plant

given that; the utility has 36% efficiency

we substitute

heat saved =  570 kWh/CLF / 36%

we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)

so

heat saved =  570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (

heat saved = 5.4 × 10⁶ Btu/CLF

i.e eat of electricity saved per CLF is 5.4 × 10⁶

b) How many 2,000-lb tons of SO₂ would not be emitted

2000 lb/tons = 5.4 × 10⁶ Btu/CLF

0.6 lb SO₂ / million Btu = x

so

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ /  million Btu )] / 2000 lb/tons

x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]

x = 3.24 × 10⁶ / 2 × 10⁹

x = 0.00162 ton/CLF

Therefore, the amount of  SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF

c)  If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?

Amount = ( SO₂ saved per CLF ) × ( rate per CFL )

we substitute

Amount = 0.00162 ton/CLF × $800

= $1.296

Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.

3 0
3 years ago
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