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2 years ago

How have the conditions under which auto fuel economy is measured changed over the years?

Engineering

1 answer:

Elina [12.6K]2 years ago

3 0

We have more technology so fuel has other factors not like before that it was more natural

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A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.

Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = $\frac{PL}{AE}$ ................1

here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = $\frac{PL}{AE}$

δ = $\frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}$

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0

3 years ago

Five hundred gallons of 89-octane gasoline is obtained by mixing 87-octane gasoline with 92-octane gasoline. (a) Write a system

miskamm [114]

Explanation:

a) The total volume equals the sum of the volumes.

500 = x + y

The total octane amount equals the sum of the octane amounts.

89(500) = 87x + 92y

44500 = 87x + 92y

b) desmos.com/calculator/ekegkzllqx

As x increases, y decreases.

c) Use substitution or elimination to solve the system of equations.

44500 = 87x + 92(500−x)

44500 = 87x + 46000 − 92x

5x = 1500

x = 300

y = 200

The required volumes are 300 gallons of 87 gasoline and 200 gallons of 92 gasoline.

6 0

3 years ago

When subject to an unknown torque, the shear stress in a 2 mm thick rectangular tube of dimension 100 mm x 200 mm was found to b

laila [671]

Answer:

The shear stress will be 80 MPa

Explanation:

Here we have;

τ = (T·r)/J

For rectangular tube, we have;

Average shear stress given as follows;

Where;

$\tau_{ave} = \frac{T}{2tA_{m}}$

$A_m$ = 100 mm × 200 mm = 20000 mm² = 0.02 m²

t = Thickness of the shaft in question = 2 mm = 0.002 m

T = Applied torque

Therefore, 50 MPa = T/(2×0.002×0.02)

T = 50 MPa × 0.00008 m³ = 4000 N·m

Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m

Therefore, $A_m$ = 0.05 m × 0.25 m = 0.0125 m².

Therefore, from the following average shear stress formula, we have;

$\tau_{ave} = \frac{T}{2tA_{m}}$

Plugging in then values, gives;

$\tau_{ave} = \frac{4000}{2\times 0.002 \times 0.0125} = 80,000,000 Pa$

The shear stress will be 80,000,000 Pa or 80 MPa.

7 0

3 years ago

Which lens is wide-angle?

PIT_PIT [208]

A wide-angle lens has a focal length of 35mm or shorter, which gives you a wide field of view. The wider your field of view, the more of the scene you'll be able to see in the frame. These lenses are ideal for many scenarios, and most photographers have at least one trusty wide-angle lens in their kit.

BRAINLIEST PLSSS

5 0

1 year ago

Two containers initially contain dry air at 40 °C. Water is added to both containers so that the relative humidity is 30% in con

enot [183]

Answer:

A) Wet bulb temperature of #1 is less than that of #2

Explanation:

This can be gotten from pinpointing the states of the two containers on a psychometric chart.

8 0

3 years ago