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Sergio [31]
3 years ago
6

What are the maximum weights for single, tandem and 5-axle combinations?

Engineering
1 answer:
belka [17]3 years ago
7 0
Hauling Vehicles that include a semitrailer manufactured prior to or in the model year of 2024, and registered in Illinois prior to January 1, 2025, having 5 axles with a distance of 42 feet or less between extreme axles, may not exceed the following maximum weights: 20,000 pounds on a single axle; 34,000 pounds
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One of the best ways to increase engine power and control detonation and preignition is to?
Reika [66]

Answer:

use water injection.

8 0
1 year ago
For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5
bonufazy [111]

Answer:

The total force resisted by the flange bolts is  163.98 N

Explanation:

Solution

The first step is to find  the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ =  diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of  the pipe at the inlet section

A₂ = π /4 D₂²

D₂ =  diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265  * 5 = 1.963 * V₂

V₂ = 12.8 m/s

Note: Kindly find an attached copy of the part of the solution to the given question below

8 0
3 years ago
Describe the engineering design process in your own words.
MrMuchimi

Answer:

Explanation:

8 0
3 years ago
Read 2 more answers
A 11-cm-diameter horizontal jet of water with a velocity of 40 m/s relative to the ground strikes a flat plate that is moving in
Reika [66]

Answer:

F = 8552.7N

Explanation:

We need first our values, that are,

V_{jet} = 40m/s\\V_{Plate} = 10m/s \\D = 11cm

We start to calculate the relative velocity, that is,

V_r = V_{jet}-V_{plate}\\V_r = (40)-(10)\\V_r = 30m/s

With the relative velocity we can calculate the mass flow rate, given by,

\dot{m}_r = \rho A V_r

\dot{m}_r = (1000)(30) \frac{\pi (0.11)^2}{4}

\dot{m}_r = 285.09kg/s

We need to define the Force in the direction of the flow,

\sum\vec{F} = \sum_{out} \beta\dot{m}\vec{V} - \sum_{in} \beta\dot{m} \vec{V}\\

F = \dot{m}V_r

F = (285.09Kg/s)(30)

F = 8552.7N

8 0
3 years ago
Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a p
oee [108]

Answer:

a) 24 kg

b) 32 kg

Explanation:

The gauge pressure is of the gas is equal to the weight of the piston divided by its area:

p = P / A

p = m * g / (π/4 * d^2)

Rearranging

p * (π/4 * d^2) = m * g

m = p * (π/4 * d^2) / g

m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg

After the weight is added the gauge pressure is 2.8kPa

The mass of piston plus addded weight is

m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg

56 - 24 = 32 kg

The mass of the added weight is 32 kg.

5 0
3 years ago
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