Question

The​ Eco-Marathon is an annual competition sponsored by Shell​ Oil, in which participants build special vehicles to achieve the highest possible fuel efficiency. The​ Eco-Marathon is held around the world with events in the United​ Kindom, Finland,​ France, Holland,​ Japan, and the United States. A world record was set in​ Eco-Marathon by a French team in 2003 called Microjoule with a performance of 10 comma 705 miles per gallon​ [mi/gal]. The Microjoule runs on ethanol. If the cars are given 100 grams​ [g] of ethanol​ (specific gravityequals0.789​) and drive until the fuel runs​ out, how far did the Microjoule drive in kilometers​ [km]?

Answer 1

Answer:

(1)  V=0.03348 gallons

(2)  L= 576.813 km .

Explanation:

Write the relation between the density and volume for ethanol.

density ( ρ )= m/V

Volume ( V )= m/ρ .......... (1)

Here, V is the volume of ethanol, m is the mass of ethanol, and  is the density of ethanol.

Write the relation between the specific gravity of ethanol and density of water

ρ=γ*ρ(water)  ............. (2)

Here,  is specific gravity of ethanol and  is the density of water.

Determine the volume of ethanol by substituting equation (2) in equation (1)

V=m/γ*ρ(water)

Substitute 100 gm , for m,  for , and 1000 kg/m  for.

V=$\frac{100*10^{-3} }{0.789*1000}$

V=0.000112674 $m^{3$

Convert m³ to gallons

V= (0.000112674*264.172 )

V=0.03348 gallons

Step 2:

Determine the distance travelled by the special vehicle named “Micro joule” runs with ethanol.

L=C*V    .......... (3)

Here,  is the rate of consumption of ethanol by Micro joule.

Substitute 10,705 mi/gallons  for C and 0.03348 gallons  for .

By using eq. 3.

L= 10750*0.03348

L=358.416 mi.

Convert the distance travelled by Micro joule from miles to kilometres

L= (358.416* 1.60934)

L = 576.813 km

Thus, the distance travelled by Micro joule is 576.813 km .

Answer 2

This question involves the concepts of density, specific gravity, and average fuel consumption.

The Microjoule drove "37.99 km" before coming to rest.

First, we will use the formula of specific gravity to calculate the density of ethanol.

$\rho_e = \gamma \rho$

where,

$\rho_e$ = density of ethanol = ?

$\gamma$ = specific gravity of ethanol = 0.789

$\rho$ = density of water = 1000 kg/m³

Therefore,

$\rho_e = (0.789)(1000\ kg/m^3)\\\rho_e = 789\ kg/m^3$

Now, we will use the density of the ethanol to find the volume of the ethanol used as fuel:

$\rho_e = \frac{m}{V}\\\\V = \frac{m}{\rho_e}$

where,

V = volume of ethanol = ?

m = mass of ethanol = 100 g = 0.1 kg

Therefore,

$V = \frac{0.1\ kg}{789\ kg/m^3}$

V = 1.26 x 10⁻⁴ m³

Now, we will convert this volume to gallons:

$V = (1.26\ x\ 10^{-4}\ m^3)(\frac{264.172\ gal}{1\ m^3})$

V = 0.0335 gal

Now, we will use the average fuel consumption to find the distance covered by Microjoule:

$Average\ Fuel\ Consumption = \frac{Distance\ Covered}{V}$

Distance Covered = D = (Average Fuel Consumption)(V)

D = (705 miles/gal)(0.0335 gal)

D = 23.6 miles

Now, we will convert this distance to kilometers (km):

D = (23.6 miles)$(\frac{1.60934\ km}{1\ miles})$

D = 37.99 km

Learn more about density here:

https://brainly.com/question/24386693?referrer=searchResults

The attached picture explains the concept of density.