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pickupchik [31]
3 years ago
7

Before you disconnect the service battery from the discharged battery, it is good practice to place a load across the

Engineering
2 answers:
Lilit [14]3 years ago
6 0

Answer:

it is true i just did this test

Explanation:

alexgriva [62]3 years ago
3 0
The answer should be true
You might be interested in
Why does an object under forced convection reach a steady-state faster than an object subjected to free-convection?
bonufazy [111]

Answer:

Free convection:

   When heat transfer occurs due to density difference between fluid then this type of heat transfer is know as free convection.The velocity of fluid is zero or we can say that fluid is not moving.

Force convection:

   When heat transfer occurs due to some external force then this type of heat transfer is know as force convection.The velocity of fluid is not zero or we can say that fluid is moving in force convection.

Heat transfer coefficient of force convection is high as compare to the natural convection.That is why heat force convection reach a steady-state faster than an object subjected to free-convection.

We know that convective heat transfer given as

 q = h  A ΔT

h=Heat transfer coefficient

A= Surface area

ΔT = Temperature difference

5 0
4 years ago
A device that converts the chemical energy from a fuel into electricity through a
Tema [17]

The answer is D, Fuel Cell.

8 0
3 years ago
Read 2 more answers
13 feet<br> 16 feet<br> 8 feet<br> 24 feet<br> + 51 feet
Allisa [31]

Answer:

112

Explanation:

because all of them add 112

6 0
3 years ago
An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin
Leni [432]

Answer:

Q=41.33 KVAR\ \\at\\\ 480 Vrms

Explanation:

From the question we are told that:

Voltage V=480/0 \textdegree V

Power P=120kW

Initial Power factor p.f_1=0.77 lagging

Final Power factor p.f_2=0.9 lagging

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

p.f_1=0.77

cos \theta_1 =0.77

\theta_1=cos^{-1}0.77

\theta_1=39.65 \textdegree

And

p.f_2=0.9

cos \theta_2 =0.9

\theta_2=cos^{-1}0.9

\theta_2=25.84 \textdegree

Therefore

Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=-41.33VAR

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

Q=41.33 KVAR\ \\at\\\ 480 Vrms

6 0
3 years ago
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
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