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const2013 [10]
3 years ago
12

1. Lea y analice la Norma ISO 16949 - Calidad en la industria automotriz, luego se ubica en los requisitos particulares, usted m

encionará de acuerdo a su criterio, cuál de los (7) requisitos, allí mencionados le parece sea el más importante o relevante para la norma. Puede ampliar un poco más generando una búsqueda de la norma por internet para conocer un poco más sobre sus requisitos para dar una respuesta acorde.
Engineering
1 answer:
Vinvika [58]3 years ago
4 0

Answer:

,,

Explanation:

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An existing building is suffering from cracks in the exterior walls. The investigating engineer wants to ensure that the foundat
jek_recluse [69]

Answer:

18 ft^{2}

Explanation:

Soil bearing pressure=\frac {Load}{Area}

Since we're given pressure of 2500 psf and load of 45000 pounds

The area=\frac {45000}{2500}=18

Therefore, the smallest area of safe footings should not be less than 18 ft^{2}

6 0
2 years ago
The mechanical advantage of a screw is always ____________________ than/to 1. Question 5 options: less, greater, equal, none of
torisob [31]

Answer:well u can use to make a shelter but that's all I can think of ??

Explanation:

3 0
3 years ago
In order to avoid a rollover, what is the highest degree incline one should mow on? 10-degree incline 5-degree incline 30-degree
ser-zykov [4K]

Answer: B: 20-degree incline

Explanation:

A tractor user should avoid slopes of more than 20 degrees in order to avoid rollovers

6 0
2 years ago
A 15-ft beam weighing 570 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground
7nadin3 [17]

Answer:

I. Tension (cable A) ≈ 6939 lbf

II. Tension (cable B) ≈ 17199 lbf

Explanation:

Let's begin by listing out the data that we were given:

mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,

g = 32.17405 ft/s²

The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.

Mathematically represented thus:

T = mg + ma

where:

T = tension, m = mass, g = gravitational force,

a = acceleration

I. For Cable A, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-20)]

T = 18339.2085 - 11400 = 6939.2085

T ≈ 6939 lbf

II. For Cable B, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-2)]

T = 18339.2085 - 1140 = 17199.2085

T ≈ 17199 lbf

4 0
3 years ago
g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,
irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where   r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where   r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0
3 years ago
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