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morpeh [17]
2 years ago
9

The specifications of an electronic device are 24 /- 0.4 Amps. When the device exceeds specifications the average quality cost i

s $32.00. A sample of the devices had an average value of 23.9 Amps and a standard deviation of 0.211 Amps. What is the average loss for this device
Engineering
1 answer:
Misha Larkins [42]2 years ago
4 0

Answer:

The average loss for this device is $10.90

Explanation:

Given data

Specification = 24 +/- 0.4 Amps

average quality cost = $32.00

average value y = 23.9

standard deviation s = 0.211 Amps

32 = k(24.4 - 24)²

32 = k(0.4)² = k(0.16)

k = 32/0.16 = 200

To evaluate average loss, use

L = k{s² + (y - T)²}

T = 24A

L = 200{0.211² + (23.9 - 24)²}

L = $10.90

The average loss for this device is $10.90

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Type the correct answer in the box. Spell all words correctly. Mike is constructing a project in which he uses a motor. How can
tankabanditka [31]
If it is. DC, direct current reverse the polarity of power leads on the motor.

If it is a 3 phase ac alternating current, reverse any of the two of three leads.

Disconnect power before attempting.
6 0
3 years ago
Steam enters a two-stage adiabatic turbine at 8 MPa and 5008C. It expands in the first stage to a state of 2 MPa and 3508C. Stea
Nataly [62]

Answer:

1) The exergy of destruction is approximately 456.93 kW

2) The reversible power output is approximately 5456.93 kW

Explanation:

1) The given parameters are;

P₁ = 8 MPa

T₁ = 500°C

From which we have;

s₁ = 6.727 kJ/(kg·K)

h₁ = 3399 kJ/kg

P₂ = 2 MPa

T₂ = 350°C

From which we have;

s₂ = 6.958 kJ/(kg·K)

h₂ = 3138 kJ/kg

P₃ = 2 MPa

T₃ = 500°C

From which we have;

s₃ = 7.434 kJ/(kg·K)

h₃ = 3468 kJ/kg

P₄ = 30 KPa

T₄ = 69.09 C (saturation temperature)

From which we have;

h₄ = h_{f4} + x₄×h_{fg} = 289.229 + 0.97*2335.32 = 2554.49 kJ/kg

s₄ =  s_{f4} + x₄×s_{fg} = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)

The exergy of destruction, \dot X_{dest}, is given as follows;

\dot X_{dest} = T₀ × \dot S_{gen} = T₀ × \dot m × (s₄ + s₂ - s₁ - s₃)

\dot X_{dest} = T₀ × \dot W×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)

∴ \dot X_{dest} = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138  - 2554.49) ≈ 456.93 kW

The exergy of destruction ≈ 456.93 kW

2) The reversible power output, \dot W_{rev} = \dot W_{} + \dot X_{dest} ≈ 5000 + 456.93 kW = 5456.93 kW

The reversible power output ≈ 5456.93 kW.

6 0
3 years ago
weight of 1000 pounds is suspended from two cables. The allowable stress in the cables is 1500 psi. Find the minimum diameter fo
kari74 [83]

Answer:

The minimum diameter for each cable should be 0.65 inches.

Explanation:

Since, the load is supported by two ropes and the allowable stress in each rope is 1500 psi. Therefore,

(1/2)(Weight/Cross Sectional Area) = Allowable Stress

Here,

Weight = 1000 lb

Cross-sectional area = πr²

where, r = minimum radius for each cable

(1/2)(1000 lb/πr²) = 1500 psi

500 lb/1500π psi = r²

r = √1.061 in²

r = 0.325 in

Now, for diameter:

Diameter = 2(radius) = 2r

Diameter = 2(0.325 in)

<u>Diameter = 0.65 in</u>

7 0
2 years ago
What is the instantaneous center of zero velocity? List two approaches for determining the is the instantaneous center of zero v
Lera25 [3.4K]

Explanation:

Instantaneous center:

   It is the center about a body moves in planer motion.The velocity of Instantaneous center is zero and Instantaneous center can be lie out side or inside the body.About this center every particle of a body rotates.

From the diagram

Where these two lines will cut then it will the I-Center.Point A and B is moving perpendicular to the point I.

If we take three link link1,link2 and link3 then I center of these three link will be in one straight line It means that they will be co-linear.

I_{12},I_{23},I_{31} all\ are\ co-linear.

5 0
3 years ago
A 10 hp motor is used to raise a 1000 Newton weight at a vertical distance of 5 meters. What is the work the motor performs?
Aleksandr [31]

The work done by a 10 HP motor when it raises a 1000 Newton weight at a vertical distance of 5 meters is <u>5kJ</u>.

Define work. Explain the rate of doing work.

Work is <u>the energy that is moved to or from an item by applying force along a displacement</u> in physics. For a constant force acting in the same direction as the motion, work is <u>easiest expressed as the product of </u><u>force </u><u>magnitude and distance traveled</u>.

Since the <u>force </u><u>transfers one unit of energy for every unit of </u><u>work </u><u>it performs</u>, the rate at which work is done and energy is used are equal.

Solution Explained:

Given,

Weight = 1000N and distance = 5m

A/Q, the work here is done in lifting then

Work = (weight) × (distance moved)

         = 1000 X 5

         = 5000Nm or 5000J = 5kJ

Therefore, the work done in lifting a 1000 Newton weight at a vertical distance of 5 meters is 5kJ.

To learn more about work, use the link given
brainly.com/question/25573309
#SPJ9

<u />

4 0
1 year ago
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