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denpristay [2]
2 years ago
7

Given the inherent costs of regulation it is safe to say that there is always a negative economic impact associated with regulat

ions.
True
False
Engineering
1 answer:
Alexus [3.1K]2 years ago
6 0

the answer is true.                                     <u>                    </u>                                                              

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Which option identifies why Ethan’s skills are valuable to his team in the following scenario?
larisa [96]

Answer:

Explanation:

The options are:

- In an isometric drawing, multiple angles and axes can be shown in one sketch.

- There is no room for detail in an isometric drawing, so the detail is shown in the orthographic projection.

- Only one sketch will be needed since all other previous designs will no longer be necessary.

- Computer programs will not be necessary to create the exact dimensions of the design.

Orthographic projections are in either the First or Third Angles but the angles are fixed and do not provide perspective view.  Isometric drawings are perspective views from different angles.

So Ethan's skill is valuable because "In an isometric drawing, multiple angles and axes can be shown in one sketch."

4 0
3 years ago
A driver is traveling at 90 km/h down a 3% grade on good, wet pavement. An accident
Paul [167]

Answer:

0.35

Explanation:

We resolve the component of the weight of the car along and perpendicular to the grade. We have mgsinФ and mgcosФ where Ф = angle of grade.

Now, the frictional force f = μN = μmgcosФ where μ = coefficient of friction

So, the net force along the grade is F = mgsinФ - μmgcosФ.

The work done by this force moving a distance, d along the grade is

W = (mgsinФ - μmgcosФ)d

This work equals the change in kinetic energy of the car. So ΔK = 1/2m(v₂² - v₁²) = W = (mgsinФ - μmgcosФ)d

1/2m(v₂² - v₁²) = (mgsinФ - μmgcosФ)d

1/2(v₂² - v₁²) = (gsinФ - μgcosФ)d

(v₂² - v₁²)/2d = (gsinФ - μgcosФ)

dividing through by gcosФ, we have

(v₂² - v₁²)/2dgcosФ = (gsinФ/gcosФ) - μgcosФ/gcosФ

(v₂² - v₁²)/2dgcosФ = tanФ -  μ

μ = tanФ - (v₂² - v₁²)/2dgcosФ

given that tanФ = 3% = 3/100 and 1 + tan²Ф = 1/cos²Ф, cosФ = 1/(√1 + tan²Ф) = 1/(√1 + (3/100)²) = 1/(√1 + (9/10000)) = 1/(√10000 + 9/10000) = 1/√(10009/10000) = 100/√10009 = 100/100.05 = 0.9995.

Also, given that v₁ = 90 km/h = 90 × 1000/3600 m/s = 25 m/s and v₂ = 45 km/h = 45 × 1000/3600 m/s = 12.5 m/s, d = 75 m and g = 9.8 m/s².

So, substituting the values of the variables into the equation, we have

μ = tanФ - (v₂² - v₁²)/2dgcosФ

μ = 3/100 - ((12.5 m/s)² - (25 m/s)²)/(2 × 75 m × 9.8 m/s² × 0.9995)

μ = 3/100 - ((156.25 m/s)² - (625 m/s)²)/1,469.265 m²/s²

μ = 3/100 - (-468.75 m²/s²)/1,469.265 m²/s²

μ = 3/100 + 468.75 m²/s²/1,469.265 m²/s²

μ = 0.03 + 0.32

μ = 0.35

So, theoretical friction  coefficient is 0.35

4 0
3 years ago
When would working with machinery be a common type of caught-in and caught-between<br> hazard?
tigry1 [53]

Answer:

A working with machinery be a common type of caught-in and caught-between  hazard is described below in complete detail.

Explanation:

“Caught in-between” accidents kill mechanics in a variety of techniques. These incorporate cave-ins and other hazards of tunneling activity; body parts extracted into unconscious machinery; reaching within the swing range of cranes and other installation material; caught between machine & fixed objects.

6 0
3 years ago
A coil with an average diameter of 5 inch will have an area of ""blank"" square meters
nadezda [96]

Answer:

19.64 square inches

Explanation:

Area will be (¶d^2)/4

= (3.142 x 5^2)/4

= 19.64 square inches

8 0
3 years ago
How do I do this?<br> Blueprints, complete the missing view.
Ymorist [56]

Explanation:

Look at the drawings and decide which view is missing. Front? Side? Top? Then draw it

7 0
3 years ago
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