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3 years ago

5

A clutch is chattering. Technician A says that uneven wear on the flywheel could be the cause. Technician B says that worn or br

oken motor mounts could be the cause. Who is correct? Group of answer choices

1 answer:

svet-max [94.6K]3 years ago

4 0

Answer:

Both Technician A and B are correct

Explanation:

Due to a contamination of the clutch disc friction surface, the clutch chatters

The contamination that causes clutch chattering includes

1) Worn or broken motor mounts

2) Bell housing bolts becoming loose

3) Clutch link damage

4) Warped flywheel, due to overheating, the flywheel can become warped such that the non uniform surface interfaces the clutch resulting in clutch chattering

5) Engine or transmission oil contaminating the disc.

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Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newton

SashulF [63]

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Explanation:

We know that for two point charges of magnitude $q_{1},q_{2}$ the magnitude of force between them is given by

$F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}$

where

$k_{e}$ is constant

$r$ is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

$F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208$

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

$F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}$

Applying value of the constant we get

$F_{1}=\frac{62.208}{0.7^{2}}$

Thus $F_{2}=126.955Newtons$

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que sabemos de la revolución industrial y como ese proceso impulso el uso de los controles eléctricos en las industrias

Alex Ar [27]

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Explanation:

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3 years ago

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stre

velikii [3]

Complete Question

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.

Answer:

The stress is $\sigma = 10. 655 MPa$

Explanation:

From the question we are told that

The critical yield resolved shear stress is $\sigma = 2.9Mpa$

First we obtain the angle $\lambda$ between the slip direction [121] and [111]

$\lambda = cos^{-1} [\frac{(u_1 u_2 + v_1 v_2 + w_1 w_2}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_2^2 + v_2^2 + w_2 ^2)} } ]$

Where $u_1 ,u_2 ,v_1 , v_2 , w_1 , w_2$ are the directional indices

$\lambda = cos ^-[ \frac{(1) (-1) + (2) (1) + (1) (1)}{\sqrt{((1)^2 +(2)^2 + (1)^2)}\sqrt{((-1)^2 + (1)^2 + (1)^2 ) } } ]$

$= cos^{-1} [\frac{2}{\sqrt{6} \sqrt{3} } ]$

$= 61.87^0$

Next is to obtain the angle $\O$ between the direction [121] and [101]

$\O = cos^{-1} [\frac{(u_1 u_3 + v_1 v_3 + w_1 w_3}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_3^2 + v_3^2 + w_3 ^2)} } ]$

Substituting 1 for $u_1$ , 2 for $v_1$ , 1 for $w_1$ , 1 for $u_2$ , 0 for $v_2$ , and 1 for $w_2$

$\O = cos^{-1} [\frac{1* 1 + 2*0 + 1*1 }{\sqrt{1^2 + 2^2 + 1^2 } \sqrt{(1^2 + 0^2 + 1^2 )} } ]$

$\O = cos^{-1} [\frac{2}{\sqrt{6} \sqrt{2} } ]$

$= 54.74 ^o$

The stress is mathematically represented as

$\sigma = \frac{\tau_c}{cos \O cos \lambda }$

$= \frac{2.9}{cos 54.74^o cos 61.87^o}$

$= \frac{2.9}{0.2722}$

$\sigma = 10. 655 MPa$

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3 years ago