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Mademuasel [1]
3 years ago
5

A student is using a 12.9 ft ramp to raise an object 6 ft above the ground.

Engineering
1 answer:
Hatshy [7]3 years ago
8 0

Answer:

2.150

Explanation:

The ideal mechanical advantage of the ramp is given as a ratio between the height of the ramp and the distance from the ground ie

Ideal mechanical advantage of the ramp=\frac {12.9}{6}=2.150

Therefore, the answer is 2.150 to the nearest thousandth

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Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase
ludmilkaskok [199]

Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite (\% V_{Gr}) is determined by the following expression:

\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%

\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%

Where:

V_{Gr} - Volume occupied by the graphite phase, measured in cubic centimeters.

V_{Fe} - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}

V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}

Where:

m_{Gr}, m_{Fe} - Masses of the graphite and ferrite phases, measured in grams.

\rho_{Gr}, \rho_{Fe} - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%

\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

m_{Gr} = \frac{2.5}{100}\times (100\,g)

m_{Gr} = 2.5\,g

m_{Fe} = 100\,g - 2.5\,g

m_{Fe} = 97.5\,g

If m_{Gr} = 2.5\,g, m_{Fe} = 97.5\,g, \rho_{Fe} = 7.9\,\frac{g}{cm^{3}} and \rho_{Gr} = 2.3\,\frac{g}{cm^{3}}, the volume percent of graphite is:

\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%

\% V_{Gr} = 91.906\,\%

The volume percent of graphite is 91.906 per cent.

6 0
2 years ago
Write down one metal or alloy that is best suited for each of the following applications:
777dan777 [17]

Answer:

Explanation:

a. Cast iron or Aluminium alloy are typically used. Aluminium is much lighter in weight and it can transfer heat better to the coolant. While Cast Iron is typically stronger and is thus still used by the manufacturers.

b. Copper can be used as a condensing heat exchanger for hot steam due to its optimal thermal properties and its ability to resist corrosion.

c. high-speed steel are perfect for producing drill bits because of its hardness and resistance to heat to an extent. Drill bits tend to produce heat as a result of the friction between it and the material to be drilled.

d. lead can be used as a container for strong acids because of its anti-corrosive properties

e.zinc and copper can be used as fuel in pyrotechnics mainly due to the fact that burn with refreshing colours. Aluminium can also be used.

f. Platinum is the metal that best suits this purpose because of its high melting point and resistivity to oxidation.

6 0
3 years ago
25 points and brainliest is it A, B, C, D
Bogdan [553]

Answer:

SIR IT IS D HOPE THIS HELPS (☞゚ヮ゚)☞☜(゚ヮ゚☜)

Explanation:

4 0
2 years ago
Read 2 more answers
Ma puteti ajuta cu un argument de 2 pagini despre inlocuirea garniturii de etansare de pe pistonul etrierului de franare la un a
Nitella [24]

Answer:

can you translate

Explanation:

what Is that?

4 0
2 years ago
Compared to 15 mph on a dry road, about how much longer will it take for
Marysya12 [62]

Answer:

8 to 10 times

Explanation:

For dry road

u= 15 mph        ( 1 mph = 0.44 m/s)

u= 6.7 m/s

Let take coefficient of friction( μ) of dry road is 0.7

So the de acceleration a = μ g

a= 0.7 x 10  m/s ²                         ( g=10 m/s ²)

a= 7 m/s ²

We know that

v= u - a t

Final speed ,v=0

0 = 6.7 - 7 x t

t= 0.95 s

For snow road

μ = 0.4

de acceleration a = μ g

a = 0.4 x 10 = 4 m/s ²

u= 30 mph= 13.41 m/s

v= u - a t

Final speed ,v=0

0 = 30 - 4 x t'

t'=7.5 s

t'=7.8 t

We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.

8 to 10 times

7 0
3 years ago
Read 2 more answers
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