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3 years ago

Describe two factors that limit the design of roller coasters

Engineering

1 answer:

serious [3.7K]3 years ago

3 0

Answer:

The factors according to the given topic are summarized below.

Explanation:

The biggest aspect becomes G forces but instead if they are handled by that of the living organism.
Coasters were already continually changing, becoming special than it's ever been and year by year they hit unprecedented levels of tremendous growth. Higher, more quickly, prolonged although innovative components have been utilized.

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How can the use of local materials improve the standard of living of Filipinos?

Rudiy27

Choose a quality one, and don't use it as necessary

7 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

The specific volume of mercury is .00007 m^3/Kg. What is its density in lbm/ft^3?

aalyn [17]

Answer:

891.027 lbm/ft³

Explanation:

See it in the pic

5 0

3 years ago

Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

3 0

3 years ago

What are the Basic requirements of drinking Water ?

iragen [17]

Answer:

1) free of contaminants, 2) alkaline, and 3) micro-clustered

Explanation:

Hope it helps you

4 0

3 years ago