The electrical panel schedules are located on EWR Plan number __

A product whose total work content time = 50 min is assembled on a manual production line at a rate of 24 units per hour. From p

valentinak56 [21]

Answer:

a)Cycle time = 2.37 min

b)Numbers of workers =21

c)Stations on the line =24

Explanation:

Given that

Total work content time(TWC) = 50 min

Production rate Rp= 24 units/hr

manning level will be close =1.5

Line balancing efficiency =0.94

a)

Cycle time

$T_c=\dfrac{60E}{R_P}$

$T_c=\dfrac{60\times 0.95}{24}$

Cycle time = 2.37 min

b)

Numbers of workers ,W

$W=\dfrac{TWC}{T_c}$

$W=\dfrac{50}{2.37}$

W= 21

Numbers of workers =21

c)

Stations on the line(n)

Lets find service time Ts

Ts = Cycle time - Time for repositioning

Ts = Tc- Tr

Ts= 2.37 - 9/ 60 min

Ts= 2.22 min

We know that efficiency

$\eta=\dfrac{TWC}{n.T_s}$

$0.94=\dfrac{50}{n\times 2.22}$

n=23.94 ⇒n=24

n=24

Stations on the line =24

7 0

3 years ago

In the ______ phase of the organizational life cycle, the organization is usually very small and agile, focusing on new products

Vlad [161]

Answer:

Entrepreneurship

Explanation:

Different phases of organizational life cycles can be observed depending on the stage of the company. These cycles are,

Entrepreneurship
Survival and Early Success
Sustained Success
Renewal (or Decline)

In the initial stage which is Entrepreneurship Phase, all of the founders take part in various activities, no formality is formed between founders and employees. Generally lots of ideas are present and company is actively searching to define correct market and products to focus on.

5 0

3 years ago

The natural material in a borrow pit has a mass unit weight of 110.0 pcf and a water content of 6%, and the specific gravity of

enot [183]

Answer:

A. 288,030.91 cy

B. The amount of water that must be removed from the natural material is 483541.04254 gallons of water

Explanation:

The natural material in the barrow properties are;

The mass unit weight, γ = 110.0 pcf

The water content, w = 6%

The specific gravity of the soil solids, $G_s$ = 2.63

The desired dry unit weight, $\gamma _d$ = 122 pcf

The water content, w₁ = 5.5 %

The net section volume, $V_T$ = 245,000 cy = 6,615,000 ft³

A. $\gamma _d$ = $W_s$ / $V_T$

∴ $W_s$ = $V_T$ × $\gamma _d$ = 6,615,000 ft³ × 122 lb/ft³ = 807030000 lbs

w = ( $W_w$ / $W_s$ ) × 100

∴ $W_w$ = (w/100) × $W_s$ = (6/100) × 807030000 lbs = 48421800 lbs

The weight of solids

W = $W_s$ + $W_w$ = 807030000 lbs + 48421800 lbs = 855451800 lbs

V = W/γ = 855451800 lbs/(110.0 lb/ft.³) = 7776834.54545 ft.³ = 288,030.91 cy

V = 288,030.91 cy

The amount of cubic yards of borrow required = 288,030.91 cy

B. The volume of water in the required soil is found as follows;

$W_{w1}$ = (w₁/100) × $W_s$ = (5.5/100) × 807030000 lbs = 44386650 lbs

The amount of water that must be added = $W_{w1}$ - $W_w$ = 44386650 lbs - 48421800 lbs = -4,035,150 lbs

Therefore, 4,035,150 lbs of water must be removed

The density of water, ρ = 8.345 lbs/gal

Therefore, V = 4,035,150 lbs/(8.345 lbs/gal) = 483541.04254 gal of water must be removed from the natural material

7 0

3 years ago

H. Blasius correlated data on turbulent friction factor in smooth pipes. His equation f s m o o t h ≈ 0.3164 Re − 1 / 4 fsmooth≈

tiny-mole [99]

Answer:

Therefore the angle the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

Explanation:

The first step to take is to calculate the the velocity of flow through a pipe

Q =Av

Where Q = is the discharge through pipe

A = Area of the pipe

v = the flow of velocity

We substitute 0.001 m^3/s for Q and 0.03 m for D

Q= Av

0.001=Av

Substitute π/4 D² for A

0.001 = π/4 D² (v)

v = 0.004/πD²

D = he diameter of the pipe

substitute 3 cm for D

v= 0.004/π * [3 cm * 1 m/100 cm]²

v =1.414 m/s

Obtain fluid properties from the table Kinematic viscosity and Dynamic of water

p =1000 kg /m³

μ= 1.002 * 10^ ⁻³ N.s/m³

Thus,

we write the expression to determine the Reynolds number of flow

Re = pvD/μ

Re = is the Reynolds number

p =density

μ = dynamic viscosity at 20⁰C

We then substitute 1000 kg /m³ in place of p, 1.002 * 10^ ⁻³ N.s/m³ for μ,

1.414 m/s for v and 0.03 m for D

Thus,

Re = 1000 * 1.414 * 0.03/ 1.002 * 10^ ⁻³ = 42335

The next step is to calculate the friction factor form the Blasius equation

f = 0.3164 (Re)^1/4

f = friction factor

We substitute 42335 for Re

f = 0.3164 (42335)1/4

=0.022

The next step is to write the expression to determine the friction head loss

hl = flv²/2gD

hl = head loss

l = length of pipe

g= acceleration due to gravity

We then again substitute 0.022 for f, 1.414 m/s for v, 0.03 m for D, and 9.8 m/s² for g.

so,

hl = flv²/2gD

hl/L = 0.022 * 1.414²/2 * 9.81 * 0.03

sinθ = 0.07473

θ = 4° 16'

Therefore the angle the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

3 0

3 years ago

Trent is designing the grounds for a massive outdoor skate park. He has no experience with skateboarding at all, and although he

alexandr1967 [171]

Answer:

I’m sorry I don’t understand this is there more steps?‍♀️

Explanation:

8 0

3 years ago

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The electrical panel schedules are located on EWR Plan number ___.