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3 years ago

Assume that Randy’s photocopying Service charges $.10 per photocopy. If fixed costs are

Engineering

1 answer:

amid [387]3 years ago

4 0

Answer:

1) 739 copies (739.7260273972603).

2) 5000 copies.

Explanation:

27000/365 = 73.97260273972603 (Their daily wins). How much times $.10 is equal to 73.97260273972603?

I used a two step equation to get it.

0.10 * x = 73.97260273972603
/0.10 = /0.10
x = 739.7260273972603

They need to sell 739.7260273972603 copies a day to get 27000$ a year.

an equation (y = mx+b) can be

y = 739.7260273972603x

To earn a 500$ profit we need to do the same two-step equation but instead of 73.97260273972603 use 500.

0.10 * x = 500
/0.10 = /0.10
x = 5000

5000 copies needs to be sold to get a 500$ profit.

The last question I don't understand but I hope those 2 questions helped

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100000000000x1000000000000=?

zhannawk [14.2K]

Answer:

Thy answer to your very sophisticated question is 1E23

Explanation:

IT JUST IS! Dont ask any questions

mAsquErade, mAsquerade tHat iS mY naME

7 0

3 years ago

Read 2 more answers

A long aluminum wire of diameter 3 mm is extruded at a temperature of 280°C. The wire is subjected to cross air flow at 20°C at

Musya8 [376]

Answer:

Explanation:

Given:

Diameter of aluminum wire, D = 3mm

Temperature of aluminum wire, $T_{s}=280^{o}C$

Temperature of air, $T_{\infinity}=20^{o}C$

Velocity of air flow $V=5.5m/s$

The film temperature is determined as:

$T_{f}=\frac{T_{s}-T_{\infinity}}{2}\\\\=\frac{280-20}{2}\\\\=150^{o}C$

from the table, properties of air at 1 atm pressure

At $T_{f}=150^{o}C$

Thermal conductivity, $K = 0.03443 W/m^oC$ ; kinematic viscosity $v=2.860 \times 10^{-5} m^2/s$ ; Prandtl number $Pr=0.70275$

The reynolds number for the flow is determined as:

$Re=\frac{VD}{v}\\\\=\frac{5.5 \times(3\times10^{-3})}{2.86\times10^{-5}}\\\\=576.92$

sice the obtained reynolds number is less than $2\times10^5$ , the flow is said to be laminar.

The nusselt number is determined from the relation given by:

$Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}$

$Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11$

The covective heat transfer coefficient is given by:

$Nu_{cyl}=\frac{hD}{k}$

Rewrite and solve for $h$

$h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K$

The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:

$Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m$

The rate of heat transfer from the wire to the air per meter length is $Q=340.42W/m$

6 0

3 years ago

A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the

Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + $\frac{u^2}{30(\frac{a}{g}\pm G)}$ ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5 + $\frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}$