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Blizzard [7]
3 years ago
9

What are three common hazardous

Engineering
1 answer:
Delicious77 [7]3 years ago
4 0

Answer:

  1. 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.
  2. COLD - Because of away from and a new cover of day off, have plunged to the youngsters and 20s. There are frigid untreated surfaces.
  3. 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.
  4. 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.

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An engineer is working with archeologists to create a realistic Roman village in a museum. The plan for a balance in a marketpla
NeTakaya

Answer:

The minimum volume requirement for the granite stones is 1543.64 cm³

Explanation:

1 granite stone weighs 10 denarium

100 granted stones will weigh 1000 denarium

1 denarium = 3.396g

1000 denarium = 3396g.

But we're told that 20% of material is lost during the making of these stones.

This means the mass calculated represents 80% of the original mass requirement, m.

80% of m = 3396

m = 3396/0.8 = 4425 g

This mass represents the minimum mass requirement for making the stones.

To now obtain the corresponding minimum volume requirement

Density = mass/volume

Volume = mass/density = 4425/2.75 = 1543.64 cm³

Hope this helps!!!

3 0
3 years ago
A large building will need several different types of workmen to install and repair pipes for water, heating,
siniylev [52]

Answer:

Plumber and pipefitters

Explanation:

3 0
3 years ago
A(n)<br> is a safety device commonly<br> used with a slotted nut.
liraira [26]

A safety device called a cotter pin. The cotter pin fits through a hole in the bolt or part. This keeps the nut from turning and possibly coming off.

5 0
3 years ago
(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th
arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation (\dot Q), measured in BTU per hour, is represented by this formula:

\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4}) (1)

Where:

\epsilon - Emissivity, dimensionless.

A - Surface area of the student, measured in square feet.

\sigma - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

T_{s} - Temperature of the student, measured in Rankine.

T_{b} - Temperature of the bus, measured in Rankine.

If we know that \epsilon = 0.90, A = 16.188\,ft^{2}, \sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}, T_{s} = 554.07\,R and T_{b} = 527.67\,R, then the heat transfer rate due to electromagnetic radiation is:

\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]

\dot Q = 417.492\,\frac{BTU}{h}

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

Q = \dot Q \cdot \Delta t (2)

Where \Delta t is the heat transfer time, measured in hours.

If we know that \dot Q = 417.492\,\frac{BTU}{h} and \Delta t = \frac{1}{3}\,h, then the net energy transfer is:

Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)

Q = 139.164\,BTU

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

7 0
3 years ago
At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
3 years ago
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