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svet-max [94.6K]
3 years ago
9

We know that passengers can be either helpful or harmful to a driver. Describe a pro and a con of having passengers in your car.

Also, explain a strategy you will use to manage teen passengers that ensures safety in your vehicle.
Engineering
2 answers:
AveGali [126]3 years ago
8 0

Question: <em>We know that passengers can be either helpful or harmful to a driver. Describe a pro and a con of having passengers in your car. Also, explain a strategy you will use to manage teen passengers that ensures safety in your vehicle.</em>

Answer: Pros:

- Pro: Most public transportation fares are cheaper than driving your car across town. There are usually discounted fares for students, disabled, seniors, and children(it may be even free to ride with a student I.D, for college students. Some buses may offer discount monthly passes as well.

- Pro: You can save gas and money, especially if your commute is long. Even if you don't get a discounted fare, the amount of money saved by not driving around could be substantial.

- Pro: There's no need to worry about being stuck in traffic, or finding a parking spot in a crowded parking lot.Whether you are headed to work or school, the added time of finding a parking spot can be the difference between being late and making it on time.  

Most buses stop near or in front of major destinations, from shopping centers to universities. This allows you to get off on or near campus without the added worry of finding parking.

- Pro: For students, the added benefit of not having to pay for a semester parking pass. The cost of semester parking permits can be the upwards of $100, so taking public transportation can eliminate that cost, and allow you to save or spend that $100 elsewhere (maybe on groceries?).

Cons:

- Con: Buses can delay, so plan on getting an hour early to your destination in preparation for delays. There are a number of reasons why a bus can be delayed, from a driver calling out sick or the bus having engine problems.    

- Con: Keeping track of the buses/trains, and the stop times can be hectic, especially in larger cities. Most timetables and transit maps are available for any major city online, but the amount of lines and routes intersecting can be overwhelming.  

Some sites offer a trip planner, where you can enter your destinations and the best route can be calculated for you. If not, there is usually time tables provided at the transit stations or on the buses themselves that you have to look at and pick out which route works best.  

- Con: Some bus stops, or areas can be unsafe, so always be aware of surroundings. Make sure your stop is well-lit at night, and always be aware of who is sitting/standing with you at a specific stop or station.  

If you don't feel comfortable waiting for a while at a particular stop, try to find an alternate stop nearby that is occupied by more people and maybe a bit safer than the original stop.

anygoal [31]3 years ago
6 0

Answer:

sub to pewdipie

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goldfiish [28.3K]

Answer:

The electrical work for the process is 256.54 Btu.

Explanation:

From the ideal gas equation:

n = PV/RT

n is the number of moles of air in the tank

P is initial pressure of air = 50 lbf/in^2 = 50 lbf/in^2 × 4.4482 N/1 lbf × (1 in/0.0254m)^2 = 344736.2 N/m^2

V is volume of the tank = 40 ft^3 = 40 ft^3 × (1 m/3.2808 ft)^3 = 1.133 m^3

T is initial temperature of air = 120 °F = (120-32)/1.8 + 273 = 321.9 K

R is gas constant = 8.314 J/mol.k

n = 344736.2×1.133/8.314×321.9 = 145.94 mol

The thermodynamic process is an isothermal process because the temperature is kept constant.

W = nRTln(P1/P2) = 145.94×8.314×321.9×ln(50/25) = 145.94×8.314×321.9×0.693 = 270669 J = 270669 J × 1 Btu/1055.06 J = 256.54 Btu

5 0
3 years ago
Read 2 more answers
A ten-station transfer machine has an ideal cycle time of 30 sec. The frequency of line stops is 0.075 stops per cycle. When a l
goldfiish [28.3K]

Answer:

62.5%

Explanation:

We are given that

A ten-station transfer machine has ideal cycle time=30 sec=\frac{30}{60}=0.5 min

Frequency of line=0.075 stops per cycle

Average time=4 min

We have to determine the line efficiency

T_p=0.5+0.075(4)=0.5+0.3=0.8

Line efficiency=\frac{0.5}{0.8}\times 100=62.5%

Hence, the line efficiency=62.5%

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Using the formula above and knowing din = 5in, dout = 40in, and nin = 20 teeth, what is nout?​
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I can’t figure nothing out with out the formula

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3 years ago
thermal energy is being added to steam at 475.8 kPa and 75% quality. determine the amount of thermal energy to be added to creat
eduard

Answer:

q_{in} = 528.6\,\frac{kJ}{kg}

Explanation:

Let assume that heating process occurs at constant pressure, the phenomenon is modelled by the use of the First Law of Thermodynamics:

q_{in} = h_{g} - h_{mix}

The specific enthalpies are:

Liquid-Vapor Mixture:

h_{mix} = 2217.2\,\frac{kJ}{kg}

Saturated Vapor:

h_{g} = 2745.8\,\frac{kJ}{kg}

The thermal energy per unit mass required to heat the steam is:

q_{in} = 2745.8\,\frac{kJ}{kg} - 2217.2\,\frac{kJ}{kg}

q_{in} = 528.6\,\frac{kJ}{kg}

7 0
4 years ago
A 0.35-ft3 well-insulated rigid can initially contains refrigerant-134a at 90 psia and 30°F. Now a crack develops in the can, an
nydimaria [60]

Answer:

m = 1.37 lbm

Explanation:

We are given that;

P1 = 90 psia

T1 = 30°F

From the table i attached, at T = 30°F, entropy, s1 = sf = 0.04752 Btu/lbm.R

We are also given;

P2 = 20 psia.

At this, s2 = s1 = 0.04752 Btu/lbm.R

From the table i attached, sf at 20 psia is; sf = 0.02605 Btu/lbm.R and sfg = 0.19962 Btu/lbm.R

Now, formula for quality of steam at Pressure P2 is;

X2 = (s2 - sf)/sfg

Plugging in the relevant values to obtain;

X2 = (0.04752 - 0.02605)/0.19962

X2 = 0.1076

Now, v2 = vf + x2•vfg

From the table i attached, at 20 psia, vf = 0.01182, vg =2.27772 and vfg = vg - vf = 2.27772 - 0.01182 = 2.2659 ft³/lbm

Thus,

v2 = 0.01182 + 0.1076*2.2659 = 0.2556 ft³/lbm

Now, let's find mass of the refrigerant from, m = V/v2

m = 0.35/0.2556 = 1.37 lbm

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3 years ago
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