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3 years ago

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Which of the following is not a reason to give yourself extra "cushion" when driving?

1 answer:

damaskus [11]3 years ago

7 0

Answer:

The question is incomplete, the complete question is:

Which of the following is not a reason to give yourself extra "cushion" when driving?

A. Poor visibility B. Poor road conditions C. Inclement weather D. None of these.

The correct answer is D. None of these.

Explanation:

All the options are not reasons to give yourself an extra cushion when driving, rather they are reasons that are not favorable to driving at all.

A cushion is a certain amount of distance you are supposed to keep between you and the car in front of you to allow easy maneuvering in any condition.

A typical cushion is 3 seconds between you and the car in front of you, in less than perfect conditions like bad weather or poor road conditions an additional second must be added to it.

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Creating vacancies in ceramics. Consider doping ZrO2 with small concentrations of Nb205. The valence of Nb is 5. Assume that Nb

Murljashka [212]

Answer:

Creating vacancies in ceramics. Consider doping ZrO₂ with small concentrations of Nb205. The valence of Nb is 5. Assume that Nb ions sit in Zr ion sites

a. A substitutional defect will be produced.

b. With this dopping, the Nb increases electron band conduction and decreases oxygen anion conduction in ZrO₂.

Explanation:

(a) The defect produced by dopping a little concentration of Nb₂O5 with Nb in the +5 charge state is known as a substitutional defect.

(b) With this dopping, the Nb increases electron band conduction and decreases oxygen anion conduction in ZrO₂.

Moreover, if oxygen vacancies are rate-limiting defect, the corrosion of ZrO₂ decreases and if electrons are rate-limiting then the corrosion of ZrO₂ is accelerated.

7 0

4 years ago

Ceramics has the weakness of resisting high compression force but low tensile force. a)-True b)-False

Nesterboy [21]

Answer: b) False

Explanation:Ceramic is brittle in nature therefore it has a tendency that it is strong during the compression and it tends to be weak during the high tensile forces. While the tensile forces are applied , ceramics are not able to yield the stress and cause breakage of the material due to high tension but does not face any fault during the compression.

7 0

4 years ago

Suppose that the cost of electrical energy is $0.15 per kilowatt hour and that your electrical bill for 30 days is $80. Assume t

mr_godi [17]

Answer:

a) 740 W b) 6.2 A c) 8.1%

Explanation:

We need first to get the total energy spent during the 30 days, that can be calculated as follows:

1 month = 30 days = 720 hr

If the total cost amounts $80 (for 720 hr), and the cost per kwh is 0.15, we have:

80 $/mo = 0.15 $/Kwh*x Kwh/mo

Solving for the total energy spent in the month:

$x (kwh/mo) = \frac{80}{0.15} = 533.3 kwh$

Assuming that the power delivered is constant over the entire 30 days, as power is the rate of change of energy, we can find the power as follows:

$P = \frac{E}{t} = \frac{533.3 kWh}{720 h} = 0.74 kW = 740 W$

b) If the power is supplied by a voltage of 120 V, we can find the current I as follows:

$I =\frac{P}{V} =\frac{740W}{120V} = 6.2A$

c) If part of the electrical load is a 60-W light, we can substract this power from the one we have just found, as follows:

P = 740 W - 60 W = 680 W

The new value of the energy spent during the entire month will be as follows:

E = 0.68 kW*24(hr/day)*30(days/mo) = 490 kWh/mo

The reduction in percentage regarding the total energy spent can be calculated as follows:

ΔE = $\frac{533.3-490}{533.3} = 0.081*100= 8.1%$

⇒ ΔE(%) = 8.1%

$%(E) =\frac{533.3-490}{533.3} = 0.081 * 100 = 8.1%$

4 0

4 years ago

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$

We can assume the same drag coeeficient $C_{D1}=C_{D2}=C_{D}$ and we got:

$\Delta P = C_{D} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D} \frac{\rho}{2}V^2 (wh) V$

$\Delta P = C_{D} \frac{\rho}{2}V^3 (A_{carrier})$

1.7 ft =0.518 m

60 mph = 26.822 m/s

In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:

$R_E= \frac{Vl}{v}= \frac{26.822m/s*0.518 m}{1.58x10^{-4} Pa s}= 8.79 x10^{4}$

And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.

Now we can find the aspect ratio like this:

$\frac{l}{h}=\frac{5}{1.7}2.941$

And we can estimate the calue of $C_D = 1.2$ from a figure.

And we can calculate the power difference like this:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

8 0

3 years ago

The name of a round stock welding position refers to the

Sedaia [141]

Charlidamelio is overrated

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3 years ago

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