Using the state mileage guide table located on the top of each map page you can

a ten station assembly machine has ideal cycle time of 6 sec. the fraction defect rate at each station 0.005 and defect always j

MAXImum [283]

Answer:

$T_{P}=(2.6667)(10^{-3})h$

Explanation:

Let's write the equation of the production rate for the assembly machine :

$T_{P}=T_{C}+(n).(m).(p).(T_{D})$

Where $T_{P}$ is the production rate for the assembly machine.

Where $T_{C}$ is the ideal cycle time

Where n is the number of stations.

Where m is the number stations that get jam when the defect occurs.

Where p is the defect rate at each station.

And where $T_{D}$ is the average downtime per breakdown

We are looking for the hourly production rate ⇒

$1h=60min\\1min=60s$ ⇒

$1h=3600s$ ⇒

$6s=\frac{(6s)(1h)}{(3600s)}= \frac{1}{600}h$

$60min=1h$ ⇒

$1.2min=\frac{(1.2min)(1h)}{(60min)}=0.02h$

$T_{P}=\frac{1}{600}h+(10)(1.0)(0.005)(0.02h)=\frac{1}{375}h=(2.6667)(10^{-3})h$

m = 1.0 in the equation.

3 0

3 years ago

Fill in the correct answer.

mestny [16]

Answer:

Explanation:

A conservator with a(n) guardianship background and knowledge of foreign languages will have more job opportunities because they broaden the scope of work a conservator can undertake.

6 0

2 years ago

P1.30 shows a gas contained in a vertical piston– cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm2 is a

iogann1982 [59]

Answer:

1.737 kJ

Explanation:

Thinking process:

Step 1

Data:

Area of the shaft = 0.8 cm²

Combined mass of shaft and piston = (24.5 + 0.5) kg

= 25 kg

Piston diameter = 0.1 m

External atmospheric pressure = 1 bar = 101.3 kPa

Pressure inside the gas cylinder = 3 bar = 3 × 101.3 kPa

g = 9.81 m/s²

Step 2

Draw a free body diagram

Step 3: calculations

area of the piston = 0.0314 m²

Change in the elevation of the piston, $\deltaz$

$\delta$ z = $\frac{PE}{mg}$

= $\frac{0.2*10x^{3} }{25*9.81}$

= 0.82 m

Next, we evaluate the work done by the shaft:

$W_{s} = F_{s} Z$

= (1668) ( 0.082)

= 1. 37 kJ

Net area for work done = A (piston) - Area of shaft

= $\pi*(0.1)^{2} - 0.8 cm^{2}$

= 77.7 cm²

= 0.007774 m²

Work done in overcoming atmospheric pressure:

Wₐ = PAZ

=101.3 kPa * 0.007774 * 0.82

= 0.637 kJ

total work = work done by shaft + work to overcome atmospheric pressure = 0.367 + 1.37

= 1.737 kJ Ans

6 0

3 years ago

What type of power source is used for gmaw

nataly862011 [7]

Answer:

The GMAW process commonly uses a constant voltage power source (GMAW-CV) that allows for a relatively constant welding voltage output over a range of welding currents. For GMAW-CV the welder selects the wire feed speed (WFS) on the wire feeder unit and an appropriate voltage on the welding power supply.

4 0

2 years ago

Read 2 more answers

Brian's kite is flying above a field at the end of 65 m of string. If the angle of elevation to the kite measures 70° , how high

tatyana61 [14]

Answer:

61.08 meters

Explanation:

To solve this question, we need to imagine a triangle, formed with the string of the kite being the hypotenuse, its projection on the ground, and the height of the kite (the result we want to calculate)

The angle formed by the ground and the kite's string is 70 degrees. The height of the triangule is opposite to that angle, and the hypotenuse is the 65 meters string.

Knowing that the height is opposite to the 70 degrees angle, we can calculate it multiplying the hypotenuse by the sine of 70 degrees, which is 0.9397:

H = 65 * sin(70) = 65 * 0.9397 = 61.08 meters

4 0

2 years ago