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goldenfox [79]
2 years ago
7

Urban cities like Atlanta have to contend with a serious problem like pollution. Drivers in California are testing out a car tha

t is fueled by hyrodgen; therefore, the emissions produced are water vapor. Read the article from the link below and answer the following questions:
goo.gl/h93L43

1. What does the author mean by the statement:
“If we’re really going to make a significant reduction in carbon emissions, you can only do that with fuel-cell vehicles in the mix"?

2. What are some of the challenges with these type of vehicles? How can these challenges be overcome?

3. Do you think fuel celled cars are a viable answer to decrease pollution why or why not?
Physics
1 answer:
riadik2000 [5.3K]2 years ago
4 0

1. With this statement, the author is referring to the fact that the vehicles are one of the largest polluters of the air. In order to reduce the pollution, the vehicles that are used will need to be changed, and with it the pollution will decrease significantly. The reduction of the pollution will come because the vehicles on hydrogen will not cause any pollution, so the enormous amounts of carbon dioxide released from the combustion of the engines will be thing of the past.

2. There are several challenges with this type of vehicles in order for them to replace the fossil fuel driven ones. The big price is one of the factors, as the majority of the people can not afford these cars. Another problem is that these vehicles are not as fast as the fossil fuel driven ones, and lot of people enjoy fast driving, despite it not being safe. There are millions of vehicles out there on the roads, and changing all of them with hydrogen vehicles will take a lot of time, as lot of those vehicles are new ones, so the people will not be willing to just throw them away and leave them rot in their garages. In order for the change of the driving park to be accomplished, the prices should go down, the people to be more serious about the environment and its protection, and patience as several decades will probably be needed for a change like this to be competed.

3. The fuel celled cars are a viable answer to decrease the pollution, as they are not causing any pollution, but instead will stop the process of large emissions of carbon dioxide from the fossil fueled cars. While this method is a good one, it should not be the only, as on its own it can not have the desirable effect, but instead all the major polluters should be included in the process. The industry and the production of energy are one of the major polluters as well, so they will need to follow the example, as if they not, the problem will stay, considering that the industry is constantly growing and the demand for energy is constantly growing too.

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A piston–cylinder device contains 0.15 kg of air initially at 2 MPa and 350°C. The air is first expanded isothermally to 500 kPa
Paraphin [41]

Answer:

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Explanation:

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3 0
2 years ago
A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th
ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

m*g=k*x

Getting the k:

k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\

The energy of a spring:

E=\frac{1}{2}*k*x^{2}

For the first case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]

For the second case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

E=\frac{1}{2}*k*x^{2}⇒k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

E=W*h=500[N]*10[m]=5000[J]

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0
2 years ago
A scientist needs to determine the average volume of five water samples collected for an experiment. What is
lorasvet [3.4K]

Answer:

D

Explanation:

ew

4 0
2 years ago
Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la
rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93  ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

v=v_0-g \,t

for the object's vertical displacement we have

y-y_0=v_0\,t - \frac{g}{2} \,t^2

If the maximum height reached by the object is given in meters, we use the value for g in m/s^2 which is: 9.8\,\,m/s^2

If the maximum height of the object is given in feet, we use the value for g in  ft/s^2  which is : 32\,\,ft/s^2

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}

and now we use this to express the maximum height in the second equation we typed:

y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g})  - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}

Then if the max height is 1.1 meters, we use the following formula to solve for v_0:

1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s

If the max height is 1.1 feet, we use the following formula to solve for v_0:

1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s

5 0
2 years ago
Read 2 more answers
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
LuckyWell [14K]

For a circular aperture, the first minima (n=1) as an angular separation from the peak of the central maxima given by

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Where,

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λ is the wavelength

λ = 545 nm = 545 × 10^-9 m

Then,

Sinθ = 1.22λ / d

Sinθ = 1.22 × 545 × 10^-9 / 4.69 × 10^-3

Sinθ = 1.418 × 10^-4 rad

Then, the head light sources have the same angular separation θ from the eye as the image have inside the eye.

For the headlight

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And let the distance of the eye be D

Then,

Sinθ = x / D

Make D subject of formula

D = x / Sinθ

D = 0.695 / 1.418 × 10^-4

D = 4902.316m

To km, 1km = 1000m

D ≈ 4.9 km

4 0
2 years ago
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