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Hatshy [7]
3 years ago
7

A disc brake rotor is being machined. Technician A says that a feed rate of 0.0008 inch per revolution and 150 RPM is where the

rotor should be set for a final cut. Technician B says that the rotor should be given a non-directional surface finish using 150 grit sandpaper. Which technician is correct?
Engineering
1 answer:
erma4kov [3.2K]3 years ago
3 0

Answer:

Technician B only

Explanation:

During rotor reconditioning, which is the process also known as machining and sanding, where sanding is the involves the application of between 120 and 150 grit sandpaper while using a non-excessive force that is applied non-directionally for up to 60 seconds on each side such that the surface roughness meets OE standards. The rotors are then cleaned by washing after they are serviced before they can then be installed.

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Finally you will implement the full Pegasos algorithm. You will be given the same feature matrix and labels array as you were gi
Diano4ka-milaya [45]

Answer:

In[7] def pegasos(feature_matrix, labels, T, L):

   """

   .

   let learning rate = 1/sqrt(t),

   where t is a counter for the number of updates performed so far       (between 1   and nT inclusive).

Args:

       feature_matrix - A numpy matrix describing the given data. Each row

           represents a single data point.

       labels - A numpy array where the kth element of the array is the

           correct classification of the kth row of the feature matrix.

       T -  the maximum number of times that you should iterate through the feature matrix before terminating the algorithm.

       L - The lamba valueto update the pegasos

   Returns: Is defined as a  tuple in which the first element is the final value of θ and the second element is the value of θ0

   """

   (nsamples, nfeatures) = feature_matrix.shape

   theta = np.zeros(nfeatures)

   theta_0 = 0

   count = 0

   for t in range(T):

       for i in get_order(nsamples):

           count += 1

           eta = 1.0 / np.sqrt(count)

           (theta, theta_0) = pegasos_single_step_update(

               feature_matrix[i], labels[i], L, eta, theta, theta_0)

   return (theta, theta_0)

In[7] (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

Out[7] (array([0.29289322, 0.29289322]), 1)

In[8] feature_matrix = np.array([[1, 1], [1, 1]])

   labels = np.array([1, 1])

   T = 1

   L = 1

   exp_res = (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

   

   pegasos(feature_matrix, labels, T, L)

Out[8] (array([0.29289322, 0.29289322]), 1.0)

Explanation:

In[7] def pegasos(feature_matrix, labels, T, L):

   """

   .

   let learning rate = 1/sqrt(t),

   where t is a counter for the number of updates performed so far       (between 1   and nT inclusive).

Args:

       feature_matrix - A numpy matrix describing the given data. Each row

           represents a single data point.

       labels - A numpy array where the kth element of the array is the

           correct classification of the kth row of the feature matrix.

       T -  the maximum number of times that you should iterate through the feature matrix before terminating the algorithm.

       L - The lamba valueto update the pegasos

   Returns: Is defined as a  tuple in which the first element is the final value of θ and the second element is the value of θ0

   """

   (nsamples, nfeatures) = feature_matrix.shape

   theta = np.zeros(nfeatures)

   theta_0 = 0

   count = 0

   for t in range(T):

       for i in get_order(nsamples):

           count += 1

           eta = 1.0 / np.sqrt(count)

           (theta, theta_0) = pegasos_single_step_update(

               feature_matrix[i], labels[i], L, eta, theta, theta_0)

   return (theta, theta_0)

In[7] (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

Out[7] (array([0.29289322, 0.29289322]), 1)

In[8] feature_matrix = np.array([[1, 1], [1, 1]])

   labels = np.array([1, 1])

   T = 1

   L = 1

   exp_res = (np.array([1-1/np.sqrt(2), 1-1/np.sqrt(2)]), 1)

   

   pegasos(feature_matrix, labels, T, L)

Out[8] (array([0.29289322, 0.29289322]), 1.0)

6 0
3 years ago
Royce has a Bevel protractor with a damaged vernier scale. What limitation will Royce face if he uses this protractor to measure
liraira [26]

Answer:

Limitation in the level of possible accuracy of the that can be obtained to mainly whole degrees

Explanation:

The angle measurement values are located around the circularly shaped bevel protractor to which a Vernier scale can be attached to increase the accuracy of the angle measurement reading

The accuracy of the bevel protractor is up to 5 arc minutes or 1/12° with the space on the Vernier scale having graduations of 1/12°, such that two spaces on the main scale is 5 arc minutes more than a space on the Vernier scale and the dimensions of the coincidence of the two scale will have an accuracy of up to 1/12°

Therefore, whereby the Vernier scale is damaged, the accuracy will be limited to the accuracy main scale reading in whole degrees.

6 0
3 years ago
Among Us Map-Skeld Imposter 3 code- DBZPTQ
Yuki888 [10]

Answer:

0:  <em>NOICE  </em>:0

Explanation:

7 0
3 years ago
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A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 19.636
luda_lava [24]

Answer:

The original length of the specimen is found to be 76.093 mm.

Explanation:

From the conservation of mass principal, we know that the volume of the specimen must remain constant. Therefore, comparing the volumes of both initial and final state as state 1 and state 2:

Initial Volume = Final Volume

πd1²L1/4 = πd2²L2/4

d1²L1 = d2²L2

L1 = d2²L2/d1²

where,

d1 = initial diameter = 19.636 mm

d2 = final diameter = 19.661 mm

L1 = Initial Length = Original Length = ?

L2 = Final Length = 75.9 mm

Therefore, using values:

L1 = (19.661 mm)²(75.9 mm)/(19.636 mm)²

<u>L1 = 76.093 mm</u>

5 0
3 years ago
Technician A says some coolant recovery systems are pressurized. Technician B says the coolant recovery system is used to keep a
Digiron [165]

Answer: Technician A

Explanation: A coolant recovery system Is one of the most important part of a vehicles cooling system. When a coolant gets too hot, it forces it way out of the spring loaded radiator cap so as to relieve pressure. Any coolant which has escaped is recovered back through the discharge tube located in the recovery tank. The system is usually air free

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