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Westkost [7]
2 years ago
9

Tech A says that as moisture levels increase in brake fluid, the boiling point of the brake fluid decreases. Tech B says that fi

eld tests by the Society of Automotive Engineers have documented that the average 1-year-old car has approximately 2% moisture in the brake fluid. Who is correct
Engineering
1 answer:
dezoksy [38]2 years ago
4 0

Answer: Both Technician A and B are correct.

Explanation:

Technicians A and B are both right about their diagnosis. The Society of Automotive Engineers performed extensive research on vehicle brake fluids and found that there is typically a 2% moisture content in the brake fluid after a year of operating a vehicle. And as the moisture content of the brake fluid rises, the boiling point of the brake fluid decreases as well.

You might be interested in
A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 34 Hz
nikitadnepr [17]

Answer:

25 - \sqrt[4]{26.66*10^{-8} }  mm

Explanation:

Given data

steel tube : outer diameter = 50-mm

power transmitted = 100 KW

frequency(f) = 34 Hz

shearing stress ≤ 60 MPa

Determine tube thickness

firstly we calculate the ; power, angular velocity and torque of the tube

power = T(torque) * w (angular velocity)

angular velocity ( w ) = 2\pif = 2 * \pi * 34 = 213.71

Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s

next we calculate the inner diameter  using the relation

  \frac{J}{c_{2}  } = \frac{T}{t_{max} }  = 467.92 / (60 * 10^6) =  7.8 * 10^-6 m^3

also

c2 = (50/2) = 25 mm

\frac{J}{c_{2} } = \frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} ) =  \frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1}  ) ]

therefore; 0.025^4 - c^{4} _{1} = 0.050 / \pi (7.8 *10^-6)

c^{4} _{1} = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)

    39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8

c_{1} = \sqrt[4]{26.66 * 10^{-8} }  =

THE TUBE THICKNESS

c_{2} - c_{1} = 25 - \sqrt[4]{26.66*10^{-8} }  mm

4 0
2 years ago
(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
2 years ago
Poles are values of Laplace transform variable, s, that make denominator of transfer function zero. Zeros are values of Laplace
Ostrovityanka [42]

Answer:

Zero 1 = -1

Zero 2 = -3

Pole 1 = 0

Pole 2 = -2

Pole 3 = -4

Pole 4 = -6

Gain = 4

Explanation:

For any given transfer function, the general form is given as

T.F = k [N(s)] ÷ [D(s)]

where k = gain of the transfer function

N(s) is the numerator polynomial of the transfer function whose roots are the zeros of the transfer function.

D(s) is the denominator polynomial of the transfer function whose roots are the poles of the transfer function.

k [N(s)] = 4s² + 16s + 12 = 4[s² + 4s + 3]

it is evident that

Gain = k = 4

N(s) = (s² + 4s + 3) = (s² + s + 3s + 3)

= s(s + 1) + 3 (s + 1) = (s + 1)(s + 3)

The zeros are -1 and -3

D(s) = s⁴ + 12s³ + 44s² + 48s

= s(s³ + 12s² + 44s + 48)

= s(s + 2)(s + 4)(s + 6)

The roots are then, 0, -2, -4 and -6.

Hope this Helps!!!

3 0
2 years ago
2An oil pump is drawing 44 kW of electric power while pumping oil withrho=860kg/m3at a rate of 0.1m3/s.The inlet and outlet diam
Natasha2012 [34]

Answer:

\eta = 91.7%

Explanation:

Determine the initial velocity

v_1 = \frac{\dot v}{A_1}

    = \frac{0.1}{\pi}{4} 0.08^2

     = 19.89 m/s

final velocity

v_2 =\frac{\dot v}{A_2}

      = \frac{0.1}{\frac{\pi}{4} 0.12^2}

      =8.84 m/s

total mechanical energy is given as

E_{mech} = \dot m (P_2v_2 -P_1v_1) + \dot m \frac{v_2^2 - v_1^2}{2}

\dot v = \dot m v                       ( v =v_1 =v_2)

E_{mech} = \dot mv (P_2 -P_1) + \dot m \frac{v_2^2 - v_1^2}{2}

                = mv\Delta P + \dot m  \frac{v_2^2 -v_1^2}{2}

                 = \dot v \Delta P  + \dot v \rho \frac{v_2^2 -v_1^2}{2}

              = 0.1\times 500 + 0.1\times 860\frac{8.84^2 -19.89^2}{2}\times \frac{1}{1000}

E_{mech} = 36.34 W

Shaft power

W = \eta_[motar} W_{elec}

    =0.9\times 44 =39.6

mechanical efficiency

\eta{pump} =\frac{ E_{mech}}{W}

=\frac{36.34}{39.6} = 0.917  = 91.7%

8 0
2 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
2 years ago
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