Best use for a suspension bridge

A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 Mpa root m is exposed to a stress of 1030 MPa. W

cupoosta [38]

Answer:

It will not experience fracture when it is exposed to a stress of 1030 MPa.

Explanation:

Given

Klc = 54.8 MPa √m

a = 0.5 mm = 0.5*10⁻³m

Y = 1.0

This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of KIc, Y, and the largest value of a in the material. This requires that we solve for σc from the following equation:

σc = KIc / (Y*√(π*a))

Thus

σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))

⇒ σc = 1382.67 MPa > 1030 MPa

Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.

3 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

Think of the differences between circuit-switching and packet-switching paradigms in the Internet core design. Assume an Interne

dem82 [27]

Answer:

0.264 ; 0.079

Explanation:

Given that:

Sample size, n = 100

Probability of being active, p = 1% = 1/100 = 0.01

Using the binomial probability relation :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Probability that more than 1 user will be active

P(x > 1) = 1 - [p(x=0) + p(x = 1)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x > 1) = 1 - [0.366 + 0.370]

P(x > 1) = 0.264

2.)

Probability that more than 2 user will be active

P(x > 2) = 1 - [p(x=0) + p(x = 1) + p(x = 2)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x = 2) = 100C2 * 0.01^2 * 0.99^98 = 0.185

P(x > 1) = 1 - [0.366 + 0.370 + 0.185]

P(x > 1) = 0.079

7 0

3 years ago

What are three common hazardous

Delicious77 [7]

Answer:

SNOW SQUALLS - Squalls are substantial snow showers that cause white-out conditions. A couple of squalls are conceivable at the beginning of today through mid-evening. The most elevated possibility is north of the expressway.
COLD - Because of away from and a new cover of day off, have plunged to the youngsters and 20s. There are frigid untreated surfaces.
WINDS – One can expect supported breezes 15-25 mph this evening, with blasts to 40 mph. Solid breezes, alongside temperatures during the 30s, will create wind chills during the 20s.
FOG - Perceivability is improving, however parts of Lancaster, Lebanon, Schuylkill, Berks, and Dauphin Provinces have decreased perceivability. With temperatures beneath freezing, mist can stick to the streets and produce frigid conditions.

Explanation:

An effect day in Pennsylvania is a day that highlights climate that will probably upset your typical every day calendar or schedule. At the point when the News 8 Tempest Group accept climate arrives at this limit, a yellow triangle with a shout imprint will show up on the figure demonstrating the climate will cause an effect. You may likewise observe the symbol utilized during a day-part figure when the News 8 Tempest Group accepts certain times may have a greater amount of an effect than others. An effect day's climate isn't viewed as outrageous or serious, perilous climate.

Light downpour, day off, ice during a morning or night drive would be instances of effect climate. Most Pennsylvania workers can deal with driving in light downpour, day off, ice, however light downpour, day off, ice at the pinnacle of busy time can upset travel or cause mishaps because of the high volume of drivers. Thick mist over the Valley which could likewise modify a drive would be delegated an effect day.

Moderate cold throughout the fall, winter, spring months, or warmth in the mid year months that may constrain open air presentation to the components would get this assignment.

The Susquehanna Valley is a significant agrarian community for the locale. Pre-summer or late-summer ice or freezes that could harm or murder vegetation would fit into this class.

Days that component consume bans from delayed droughts combined with solid breeze blasts likewise would be considered to conceivably make an effect a day ahead.

4 0

3 years ago

How can you avoid aliasing error during sampling when converting an analog signal?

Romashka-Z-Leto [24]

Answer and Explanation:

Aliasing is a distortion in the signal when a continuous signal is converted into digital signal that is ADC process, due to this effect the signal aliases of one another.There will be no aliasing effect if the frequency of the signal will not be higher than the sampling frequency

The another way of avoiding aliasing is to limit the range of the continuous signal.

7 0

3 years ago