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I am Lyosha [343]

3 years ago

13

A European car manufacturer reports that the fuel efficiency of the new MicroCar is 48.5 km/L highway and 42.0 km/L city. What a

re the equivalent fuel efficiency rates in miles per gallon?

1 answer:

statuscvo [17]3 years ago

3 0

Answer:

Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon

Explanation:

1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre = 183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon

183.5919 km/gallon = 114.08 miles/gallon (highway)

158.9868 km/gallon = (158.9868 /1.6093) miles/gallon

158.9868 km/gallon = 98.79 miles/gallon (city)

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Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of "84" m3/min and exits at 12

Nina [5.8K]

Answer:

W = - 184.8 kW

Explanation:

Given data:

$P_1 = 1.05 bar$

$T_1 = 300K$

$\dot V_1 = 84 m^3/min$

$P_2 = 12 bar$

$T_2 = 400 K$

We know that work is done as

$W = - [ Q + \dor m[h_2 - h_1]]$

for $P_1 = 1.05 bar, T_1 = 300K$

density of air is 1.22 kg/m^3 and $h_1 = 300 kJ/kg$

for $P_2 = 12 bar, T_2 = 400 K$

$h_2 = 400 kj/kg$

$\dot m = \rho \times \dor v_1 = 1.22 \frac{84}{60} =1.708 kg/s$

W = -[14 + 1.708[400-300]]

W = - 184.8 kW

8 0

3 years ago

For a fluid with a Prandtl Number of 1000.0, the hydrodynamic layer is thinner than the thermal boundary layers. a) True b) Fals

kvv77 [185]

Answer:

(b)False

Explanation:

Given:

Prandtl number(Pr) =1000.

We know that $Pr=\dfrac{\nu }{\alpha }$

Where $\nu$ is the molecular diffusivity of momentum

$\alpha$ is the molecular diffusivity of heat.

Prandtl number(Pr) can also be defined as

$Pr=\left (\dfrac{\delta }{\delta _t}\right )^3$

Where $\delta$ is the hydrodynamic boundary layer thickness and $\delta_t$ is the thermal boundary layer thickness.

So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

In given question Pr>1 so hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

So hydrodynamic layer will be thicker than the thermal boundary layer.

8 0

4 years ago

(I really need help ASAP please!! this is for science her is the problem)

grandymaker [24]

Answer:

Explanation:

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5 0

3 years ago

Read 2 more answers

What is the magnitude of the maximum stress that exist at the tip of an internal crack having a radius of curvature of 1.9 x 10-

Hitman42 [59]

Answer:

2800 [MPa]

Explanation:

In fracture mechanics, whenever a crack has the shape of a hole, and the stress is perpendicular to the orientation of such, we can use a simple formula to calculate the maximum stress at the crack tip

$\sigma_{m} = 2 \sigma_{p} (\frac{l_{c}}{r_{c}})^{0.5}$

Where $\sigma_{m}$ is the magnitude of he maximum stress at the tip of the crack, $\sigma_{p}$ is the magnitude of the tensile stress, $l_{c}$ is $1/2$ the length of the internal crack, and $r_{c}$ is the radius of curvature of the crack.

We have:

$r_{c}=1.9*10^{-4} [mm]$

$l_{c}=3.8*10^{-2} [mm]$

$\sigma_{c}=140 [MPa]$

We replace:

$\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}$

We get:

$\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}=2800 [MPa]$

5 0

3 years ago

A metal shear can be used to cut flat stock , round stock , channel iron and which of the following?

Charra [1.4K]

Answer:

Angle Iron

Explanation:

Mark brainliest please.

3 0

3 years ago