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4 years ago

What are the 2 reasons an alignment should be done?

Engineering

1 answer:

NikAS [45]4 years ago

5 0

Answer:

because it will keep the tires in much better shape and it can improve the handling and keep your car from pulling to one side

Explanation:

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You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210

Naily [24]

Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

<u>Calculate the kinematic viscosity of Fluid A </u>

First step : determine the density of fluid A

Pa = Pw * Specific gravity = 1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

= 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

4 0

3 years ago

What is a combination circuit? A combination circuit:

Anon25 [30]

Answer:

Combination circuit; The basic strategy for the analysis of combination circuits involves using the meaning of equivalent resistance for parallel branches to transform the combination circuit into a series circuit.

Example:

The use of both series and parallel connections within the same circuit. In this case, light bulbs A and B are connected by parallel connections and light bulbs C and D are connected by series connections. This is an example of a combination circuit.

7 0

3 years ago

Fully developed conditions are known to exist for water flowing through a 25-mm-diameter tube at 0.01 kg/s and 27 C. What is the

Irina18 [472]

Answer:

0.0406 m/s

Explanation:

Given:

Diameter of the tube, D = 25 mm = 0.025 m

cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²

Mass flow rate = 0.01 kg/s

Now,

the mass flow rate is given as:

mass flow rate = ρAV

where,

ρ is the density of the water = 1000 kg/m³

A is the area of cross-section of the pipe

V is the average velocity through the pipe

thus,

0.01 = 1000 × 4.9 × 10⁻⁴ × V

V = 0.0203 m/s

also,

Reynold's number, Re = $\frac{VD}{\nu}$

where,

ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s

thus,

Re = $\frac{0.0203\times0.025}{0.833\times10^{-6}}$

Re = 611.39 < 2000

thus,

the flow is laminar

hence,

the maximum velocity = 2 × average velocity = 2 × 0.0203 m/s

maximum velocity = 0.0406 m/s

5 0

3 years ago

What type of oil pressure gauge should be used when

liq [111]

Answer:

The mechanical gauge would be the one for the job

Explanation:

6 0

3 years ago

An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at

polet [3.4K]

Answer:

q = 1.73 W

Explanation:

given data

small end = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

thermal conductivity of asbestos = 0.173 W/mK

solution

first we will get here side of cross section that is express as

$S = S1 + \frac{S2-S1}{L} x$ ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + $\frac{10-5}{15} x$

S = $\frac{0.15 + x}{3}$ m

and

now we get here Area of section at distance x is

area A = S² ...............2

area A = $(\frac{0.15 + x}{3})^2$ m²

and

now we take here small length dx and temperature difference is dt

so as per fourier law

heat conduction is express as

heat conduction q = $\frac{-k\times A\ dt}{dx}$ ...............3

put here value and we get

heat conduction q = $-k\times (\frac{0.15 + x}{3})^2 \ \frac{dt}{dx}$

it will be express as

$q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)$

now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K

$q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}$

solve it and we get

q (30) = (0.173) × (600 - 300)

q = 1.73 W

5 0

3 years ago