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2 years ago

For automobile engines, identify the best choice of motor oil.

Engineering

1 answer:

Vlad1618 [11]2 years ago

6 0

Answer:

C. Low viscosity for Maine in the winter and high viscosity for Florida in the summer.

Explanation:

The choice of motor oil or engine oil depends on the type of engine and also on the type of climate on which the engine runs.

In winter climates, we should use oil with low viscosity as at lower temperatures, the it flows more readily if its viscosity is low.

But the summer season needs a relatively thicker oil so as to reduce the thinning effect of the oil out of the engine.

Thus for Maine in the winter season, we need a low viscosity oil and for Florida a high viscosity oil in the summer season.

Therefore, option (c) is correct.

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A manufacturer has been asked to produce 100 customized metal discs with a particular pattern engraved on them. Which production

topjm [15]

I’m just here for points because I have test and I need them lol

8 0

2 years ago

Read 2 more answers

A heat engine operates between a source at 477°C and a sink at 27°C. If heat is supplied to the heat engine at a steady rate of

lara [203]

Answer:

$T_C = 27+273.15 = 300.15 K$

$T_H = 477+273.15 = 750.15 K$

And replacing in the Carnot efficiency we got:

$e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%$

$W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}$

Explanation:

For this case we can use the fact that the maximum thermal efficiency for a heat engine between two temperatures are given by the Carnot efficiency:

$e = 1 -frac{T_C}{T_H}$

We have on this case after convert the temperatures in kelvin this:

$T_C = 27+273.15 = 300.15 K$

$T_H = 477+273.15 = 750.15 K$

And replacing in the Carnot efficiency we got:

$e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%$

And the maximum power output on this case would be defined as:

$W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}$

Where $Q_H$ represent the heat associated to the deposit with higher temperature.

4 0

2 years ago

John has an exhaust leak in his Acura Integra GS-R, What steps would he take to fix the leak in time for his inspection?

SCORPION-xisa [38]

Answer:

Explanation:

Fist you need to identify where the leak is coming from. You can do this by either listening for the leak or spraying soapy water on the exhaust to look for air bubbles coming out of the exhaust. Depending on the spot of the leak there are many ways you can fix this leak.

1. Exhaust clamp

2. Exhaust putty

3. Exhaust tape

4. New exhaust

Exhaust clamp is best used for holes on straight pipes.

Putty is best used on welds or small holes like on exhaust manifolds or welds connecting various pieces like catalytic converters, mufflers, or resonators.

Tape will work best on straight pipes with holes.

New exhaust is for when the thig is beyond repair, like rust.

Now good luck because working on exhausts is a pain.

4 0

1 year ago

An astronomer of 65 kg of mass hikes from the beach to the observatory atop the mountain in Mauna Kea, Hawaii (altitude of 4205

lara [203]

Answer:

$0.845\ \text{N}$

Explanation:

g = Acceleration due to gravity at sea level = $9.81\ \text{m/s}^2$

R = Radius of Earth = 6371000 m

h = Altitude of observatory = 4205 m

Change in acceleration due to gravity due to change in altitude is given by

$g_h=g(1+\dfrac{h}{R})^{-2}\\\Rightarrow g_h=9.81\times(1+\dfrac{4205}{6371000})^{-2}\\\Rightarrow g_h=9.797\ \text{m/s}^2$

Weight at sea level

$W=mg\\\Rightarrow W=65\times 9.81\\\Rightarrow W=637.65\ \text{N}$

Weight at the given height

$W_h=mg_h\\\Rightarrow W_h=65\times 9.797\\\Rightarrow W_h=636.805\ \text{N}$

Change in weight $W_h-W=636.805-637.65=-0.845\ \text{N}$

Her weight reduces by $0.845\ \text{N}$ .

8 0

2 years ago

The diffusion coefficients for species A in metal B are given at two temperatures:

Kruka [31]

Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

Diffusion is governed by Arrhenius equation

$D = D_0e^{\frac{-Q_d}{RT} }$

I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

$R = 8.314\ J/mol*K$

and

°C + 273 = K

here, adjust your precision as neccessary

Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

$ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}$