Introduce JTA and JT

Currently, the lost time of each stage is 4 seconds, and intersection critical v/c ratio is set to be 0.75 to avoid cycle failur

KIM [24]

Find the solution in the attachments

Note: Question was incomplete, so the complete question is added in the attachments.

7 0

3 years ago

Technician A says that the neutral safety switch must be bad and should be replaced. Technician B says that the neutral safety s

lukranit [14]

Answer:

Technician A is correct

Explanation:

The neutral safety switch when bad must be replaced and not adjusted as suggested by technician B because if the neutral safety switch is bad the Engine might not crank when put in neutral but it will crank when put in park and this is very bad for the life of the Engine it is better to replace it. A test for a bad/faulty neutral safety switch will be required to ascertain the level of damage it might cause to the Engine and prompt replacement is essential as well. because the neutral helps to prevent the car from starting when it is already engaged in a gear position therefore protecting the car from sudden collisions

3 0

3 years ago

Read 2 more answers

Consider a mixing tank with a volume of 4 m3. Glycerinflows into a mixing tank through pipe A with an average velocity of 6 m/s,

Svetach [21]

This question is incomplete, the complete question as well as the missing diagram is uploaded below;

Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.

Take $p_o$ = 880 kg/m³ and $p_{glycerol$ = 1260 kg/m³

Answer:

the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³

Explanation:

Given that;

Inlet velocity of Glycerin, $V_A$ = 6 m/s

Inlet velocity of oil, $V_B$ = 3 m/s

Density velocity of glycerin, $p_{glycerol$ = 1260 kg/m³

Density velocity of glycerin, Take $p_o$ = 880 kg/m³

Volume of tank V = 4 m

from the diagram;

Diameter of glycerin pipe, $d_A$ = 100 mm = 0.1 m

Diameter of oil pipe, $d_B$ = 80 mm = 0.08 m

Diameter of outlet pipe $d_C$ = 120 mm = 0.12 m

Now, Appling the discharge flow equation;

$Q_A + Q_B = Q_C$

$A_Av_A + A_Bv_B = A_Cv_C$

π/4 × ( $d_A$ )² $v_A$ + π/4 × ( $d_B$ )² $v_B$ = π/4 × ( $d_C$ )² $v_C$

we substitute

π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)² $v_C$

0.04712 + 0.0150796 = 0.0113097 $v_C$

0.0621996 = 0.0113097 $v_C$

$v_C$ = 0.0621996 / 0.0113097

$v_C$ = 5.5 m/s

Now we apply the mass flow rate condition

$m_A + m_B = m_C$

$p_{glycerin}A_Av_A + p_0A_Bv_B = pA_Cv_C$

so we substitute

1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5

1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335

59.3712 + 13.27 = 0.06220335p

72.6412 = 0.06220335p

p = 72.6412 / 0.06220335

p = 1167.8 kg/m³

Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³

4 0

3 years ago

The velocity components u and v in a two-dimensional flow field are given by: u = 4yt ft./s, v = 4xt ft./s, where t is time. Wha

zmey [24]

Answer:

vec(a) = 16 i + 16 j

mag(a) = 22.63 ft/s^2

Explanation:

Given,

- The two components of velocity are given for fluid flow:

u = 4*y ft/s

v = 4*x ft/s

Find:

What is the time rate of change of the velocity vector V (i.e., the acceleration vector) for a fluid particle at x = 1 ft. and y = 1 ft. at time t = 1 second?

Solution:

- The rate of change of velocity is given to be acceleration. We will take derivative of each components of velocity with respect to time t:

a_x = du / dt

a_x = 4*dy/dt

a_y = dv/dt

a_y = 4*dx/dt

- The expressions dx/dt is the velocity component u and dy/dt is the velocity component v:

a_x = 4*(4*y) = 16y

a_y = 4*(4*x) = 16x

- The acceleration vector can be expressed by:

vec(a) = 16y i + 16x j

- Evaluate vector (a) at x = 1 and y = 1:

vec(a) = 16*1 i + 16*1 j = 16 i + 16 j

- The magnitude of acceleration is given by:

mag(a) = sqrt ( a^2_x + a^2_y )

mag(a) = sqrt ( 16^2 + 16^2 )

mag(a) = 22.63 ft/s^2

7 0

3 years ago

*13–52. A girl, having a mass of 15 kg, sits motionless relative to the surface of a horizontal platform at a distance of r = 5

Dennis_Churaev [7]

Answer:

w = 0.626 rad / s, v = 3.13 m/s

Explanation:

For this exercise let's use Newton's second law

F = m a

Where the force is a friction force and the acceleration is centripetal,

a = v² / r = w² r

The formula for friction force

fr = μ N

In a free body diagram

N- W = 0

W = N

The frictiμon outside goes from zero to the maximum value, let's calculate the speed for the maximum value of the friction force, replace

μ m g = m w² r

w = √ μ g / r

Let's calculate

w = √(0.2 9.8 / 5)

w = 0.626 rad / s

angular and linear velocity are related

v = w r

v = 0.626 5

v = 3.13 m/s

6 0

3 years ago