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3 years ago

11

A bicycle has tires that are 26 inches in diameter. The while in motion, the tires rotate through 125 complete revolutions. How

fast has the bicycle traveled? Give the answer in feet

1 answer:

Stolb23 [73]3 years ago

3 0

Answer:

850.8480103 feet

Explanation:

First you take the diameter and find the circumference, which is (2)(pi)(r) plug in your r which is 26/2= 13 so 2(13)(pi) and multiply taht by 125 after that take your answer and divide by 12which is 850.8480103

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The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago

Consider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 k

In-s [12.5K]

Consider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pmConsider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pmConsider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pm

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3 0

3 years ago

The first thing you should do is develop a ____________________ to determine what vehicle you can afford.

swat32

The first thing you should do is develop a <u>budget</u> to determine what vehicle you can afford.

<h3>What is an automobile?</h3>

An automobile is also referred to as a vehicle, car or motorcar and it can be defined as a four-wheeled vehicle that is designed and developed to be propelled by an internal-combustion (gasoline) engine, especially for the purpose of transportation from one location to another.

<h3>What is a budget?</h3>

A budget can be defined as a financial plan that is typically used for the estimation of revenue and expenditures of an individual, business organization or government for a specified period of time, often one year.

In this context, we can reasonably infer and logically deduce that the first thing anyone should do is to develop a <u>budget</u> in order to determine what vehicle they can afford.

Read more on budget here: brainly.com/question/13964173

#SPJ1

7 0

1 year ago

The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm

aksik [14]

Answer:

Tmax=14.5MPa

Tmin=10.3MPa

Explanation:

T = 600 * 0.15 = 90N.m

$T_max =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}$

=14.5MPa

$T_{min} =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}$

=10.3MPa

7 0

3 years ago

I will mark as brainliest !

Sliva [168]

Answer:

7.8 Mph

Explanation:

Rate of cycling = 1.1 rev/s

Rear wheel diameter = 26 inches

Diameter of sprocket on pedal = 6 inches

Diameter of sprocket on rear wheel = 4 inches

Circumference of rear wheel = \pi d=26\piπd=26π

Speed would be

\begin{gathered}\text{Rate of cycling}\times \frac{\text{Diameter of sprocket on pedal}}{\text{Diameter of sprocket on rear wheel}}\times{\text{Circumference of rear wheel}}\\ =1.1\times \frac{6}{4}\times 26\pi\\ =134.77432\ inches/s\end{gathered}Rate of cycling×Diameter of sprocket on rear wheelDiameter of sprocket on pedal×Circumference of rear wheel=1.1×46×26π=134.77432 inches/s

Converting to mph

1\ inch/s=\frac{1}{63360}\times 3600\ mph1 inch/s=633601×3600 mph

134.77432\ inches/s=134.77432\times \frac{1}{63360}\times 3600\ mph=7.65763\ mph134.77432 inches/s=134.77432×633601×3600 mph=7.65763 mph

The Speed of the bicycle is 7.8 mph

3 0

3 years ago