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3 years ago

What must you do if hauling a load of material which could fall or blow onto the roadway?

1 answer:

7 0

Most likely, what you must do is secure the load that you want to carry. This is extremely important if you want to haul a load of garbage in the back of your car. Securing your load properly will ensure that none of it ends up on the road, polluting the environment. This will also ensure that the garbage does not fall off the truck and hit other cars or drivers.

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Which of the following maintenance items helps to ensure the vehicles engine lasts as long as possible?

Paladinen [302]

Answer:

because oil is preventing corrosion and rust

8 0

2 years ago

The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content

MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

$Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf$

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

$\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}$

Therefore, $V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}$

$V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy$

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since $1 yd^{3}= 27 ft^{3}$

The embankment requires water of 6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

$Unit_{weight}=\frac {17451720}{498594}=35.00186 lb$

1 gallon is approximately $8.35 yd^{3}$ hence

$\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}$

That's approximately 4.2 gallons

7 0

3 years ago

Why do you suppose a value of 5 is used? Do you think other values might work?

Shtirlitz [24]

Answer:

YES

Explanation:

values other than five will work

3 0

2 years ago

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m

zvonat [6]

Answer:

T = 858.25 s

Explanation:

Given data:

Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, c 550 J/kg K, k 48 W/m K),

initial uniform temperature ( Ti ) = 200 c

Final temperature = 550 c

convection coefficient = 250 w/m^2 k

products combustion temp = 800 c

calculate how long the plate should be left in the furnace ( to attain 550 c )

first calculate/determine the Fourier series Number ( Fo )

$\frac{T_{0}-T_{x} }{T_{1}-T_{x} } = C_{1} e^{(-0.4888^{2}*Fo )}$

= 0.4167 = $1.0396e^{-0.4888*Fo}$

therefore Fo = 3.8264

Now determine how long the plate should be left in the furnace

Fo = $(\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )$

k = 48

p = 7830

L = 0.1

Input the values into the relation and make t subject of the formula

hence t = 858.25 s

8 0

3 years ago

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure

and we know that is also in term of change in density also

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

here ρ is density

and we know ρ for 20°C = 998 kg/m³

and ρ for 80°C = 972 kg/m³

so from equation 2 put all value

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

so now % change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

put here all value

0.0130 = $\frac{\pi }{4} *2^2*dl$

dl = 0.004138 m

so water level rise is 4.138 mm

8 0

3 years ago