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4 years ago

14

Two production methods are being compared. One manual and the other automated. The manual method produces 10 pc per hour and req

uires one worker at $ 15.00 per hour. Fixed cost of the manual method is $ 5,000 per year. The automated method produces 25 pc per hour, has a fixed cost of $ 55,000 per year, and a variable cost of $ 4.50 per hour. Determine the Break – Even Point (BEP) for the two methods; that is, determine the annual production quantity at which the two methods have the same annual cost. Ignore the cost of material used in the two methods

1 answer:

Genrish500 [490]4 years ago

6 0

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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. Using the Newton Raphson method, determine the uniform flow depth in a trapezoidal channel with a bottom width of 3.0 m and si

Over [174]

Answer:

y ≈ 2.5

Explanation:

Given data:

bottom width is 3 m

side slope is 1:2

discharge is 10 m^3/s

slope is 0.004

manning roughness coefficient is 0.015

manning equation is written as

$v =1/n R^{2/3} s^{1/2}$

where R is hydraulic radius

S = bed slope

$Q = Av =A 1/n R^{2/3} s^{1/2}$

$A = 1/2 \times (B+B+4y) \times y =(B+2y) y$

$R =\frac{A}{P}$

P is perimeter $= (B+2\sqrt{5} y)$

$R =\frac{(3+2y) y}{(3+2\sqrt{5} y)}$

$Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}$

solving for y $100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}$

solving for y value by using iteration method ,we get

y ≈ 2.5

5 0

3 years ago

A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. A

OLEGan [10]

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

<u>Determine the maximum speed for safe vehicle operation </u>

firstly calculate the stopping sight distance

m = R ( 1 - cos $\frac{28.655*S}{R}$ ) ---- ( 1 )

R = 678

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) ) = 30 / 678 = 0.044

cos $\frac{28.65 *s }{678}$ = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut + $\frac{u^2}{30(\frac{a}{3.2} )-G1}$ ---- ( 2 )

t = reaction time, a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + $\frac{u^2}{30[(11.2/32.2)-0 ]}$

3.675 u + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

5 0

3 years ago

In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water

Orlov [11]

Answer:

The amount of energy transferred to the water is 4.214 J

Explanation:

The given parameters are;

The mass of the object that drops = 5 kg

The height from which it drops = 86 mm (0.086 m)

The potential energy P.E. is given by the following formula

P.E = m·g·h

Where;

m = The mass of the object = 5 kg

g = The acceleration de to gravity = 9.8 m/s²

h = The height from which the object is dropped = 0.086 m

Therefore;

P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J

Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;

The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.

6 0

3 years ago

Calculate the viscosity(dynamic) and kinematic viscosity of airwhen

nikitadnepr [17]

Answer:

(a) dynamic viscosity = $1.812\times 10^{-5}Pa-sec$

(b) kinematic viscosity = $1.4732\times 10^{-5}m^2/sec$

Explanation:

We have given temperature T = 288.15 K

Density $d=1.23kg/m^3$

According to Sutherland's Formula dynamic viscosity is given by

${\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}$ , here

μ = dynamic viscosity in (Pa·s) at input temperature T,

$\mu _0$ = reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

$T_0$ = reference temperature in kelvin,

C = Sutherland's constant for the gaseous material in question here C =120

$\mu _0=4\pi \times 10^{-7}$

$T_0$ = 291.15

$\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-s$ when T = 288.15 K

For kinematic viscosity :

$\nu = \frac {\mu} {\rho}$

$kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec$

3 0

3 years ago

A vertical pole consisting of a circular tube of outer diameter 127 mm and inner diameter 115 mm is loaded by a linearly varying

Anna [14]

Maximum shear stress in the pole is 0.

<u>Explanation:</u>

Given-

Outer diameter = 127 mm

Outer radius, $r_{2}$ = 127/2 = 63.5 mm

Inner diameter = 115 mm

Inner radius, $r_{1}$ = 115/2 = 57.5 mm

Force, q = 0

Maximum shear stress, τmax = ?

τmax $= \frac{4q}{3\pi } (\frac{r2^2 + r2r1 + r1^2}{r2^4 - r1^4} )$

If force, q is 0 then τmax is also equal to 0.

Therefore, maximum shear stress in the pole is 0.

3 0

4 years ago