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2 years ago

Technician A says automotive tires are pneumatic, Technician B says modern automotive tires are not

Engineering

1 answer:

vekshin12 years ago

8 0

Answer:

Explanation:

Most tires, such as those for automobiles and bicycles, are pneumatically inflated structures, which also provide a flexible cushion that absorbs shock as the tire rolls over rough features on the surface.

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The rigid bar CDE is attached to a pin support at E and rests on the 30 mm diameter brass cylinder BD. A 22 mm diameter steel ro

Leno4ka [110]

Answer:

stress = 38.84 MPa

Explanation:

$S_{D} = \alpha _{brass} * (delta T) *(L_{BD} )\\= (18.8 * 10^(-6) )*(30)*(0.3)\\= 0.0001692 m\\\\E_{BD} = stress / strain\\stress = E_{BD} * S_{D}\\stress = (200 *10^9) * (0.0001692)\\stress = 33.84 MPa$

3 0

2 years ago

Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 u

Licemer1 [7]

Answer:

the difference in pressure between the inside and outside of the droplets is 538 Pa

Explanation:

given data

temperature = 68 °F

average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

σ = 2.69 × $10^{-2}$ N/m

so average radius = $\frac{diameter}{2}$ = 100 µm = 100 × $10^{-6}$ m

now here we know relation between pressure difference and surface tension

so we can derive difference pressure as

2π×σ×r = Δp×π×r² .....................1

here r is radius and Δp pressure difference and σ surface tension

Δp = $\frac{2 \sigma }{r}$

put here value

Δp = $\frac{2*2.69*10^{-2}}{100*10^{-6}}$

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

7 0

2 years ago

Use Lagrange multiplier techniques to find the local extreme values of the given function subject to the stated constraint. If a

Leno4ka [110]

Answer:

Explanation:

Given f(x, y) = 5x + y + 2 and g(x, y) = xy = 1

The step by step calculation and appropriate substitution is clearly shown in the attached file.

5 0

2 years ago

Problem 5) Water is pumped through a 60 m long, 0.3 m diameter pipe from a lower reservoir to a higher reservoir whose surface i

kap26 [50]

Answer:

\epsilon = 0.028*0.3 = 0.0084

Explanation:

\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2

where P_1 = P_2 = 0

V1 AND V2 =0

Z1 =0

h_P = \frac{w_p}{\rho Q}

=\frac{40}{9.8*10^3*0.2} = 20.4 m

20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10

we know thaTV =\frac{Q}{A}

V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec

20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10

f = 0.0560

Re =\frac{\rho v D}{\mu}

Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5

fro Re = 7.53*10^5 and f = 0.0560

\frac{\epsilon}{D] = 0.028

\epsilon = 0.028*0.3 = 0.0084

4 0

2 years ago

Determine the design angle ϕ (0∘≤ϕ≤90 ∘) between struts AB and AC so that the 400 lb horizontal force has a component of 600 lb

Svetllana [295]

Answer:

design angle ∅ = 4.9968 ≈ 5⁰

Explanation:

First calculate the force Fac :

Fac = $\sqrt{400^2 + 650^2 - 2(400)(650)cos30}$

= $\sqrt{160000 + 422500 - 80210}$

= 708.72 Ib

using the sine law to determine the design angle

$\frac{sin}{400} = \frac{sin 30}{Fac}$

hence ∅ = $sin^{-1} (\frac{sin 30 *400}{708.72} )$

= $sin^{-1} 0.0871$ = 4.9968 ≈ 5⁰

4 0

2 years ago