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lesantik [10]
3 years ago
11

Three consequences of unbalanced wheels on a motor vehicle​

Engineering
1 answer:
nalin [4]3 years ago
8 0
Unstable car and loss of grip on the road. A hazard to other vehicles and pedestrians as well as a very shaky car and risk of your wheels rolling off and riding on rims.
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Find all the words, Figure out my puzzle!
Mandarinka [93]

Answer:

I found the word!

It was Free points.

5 0
3 years ago
Read 2 more answers
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)}

and

ln(6.56*10^{-16}) = ln(D_0) -\frac{Q_d}{R*(1290+273)}

You might notice that these equations have the form of  

d=y-ax

You can solve this equation system easily using calculator, and you will eventually get

D_0 =6.11*10^{-11}\ m^2/s\\ Q_d=1.49 *10^3\ J/mol

After you got those 2 parameters, the rest is easy, you can just plug them all   including the given temperature of 1180°C into the Arrhenius equation

6.11*10^{-11}e^{\frac{149\ 000}{8.143*(1180+273)}

And you should get D = 2.76*10^-16 m^/s as an answer for c)

5 0
3 years ago
A vacuum pump is used to drain a basement of 20 °C water (with a density of 998 kg/m3 ). The vapor pressure of water at this tem
lord [1]

Answer:

The maximum theoretical height that the pump can be placed above liquid level is \Delta h=9.975\,m

Explanation:

To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature.  As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:

\frac{\Delta P}{\rho}+g\, \Delta h +\frac{1}{2}  \Delta v^2 =0

(\rho stands here for density, h for height)

Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:

\frac{\Delta P}{\rho}+g\, \Delta h  =0

\Delta P= -g\, \rho\, \Delta h

This means that pressure drop is proportional to the suction lift's height.

We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.

That means:

\Delta P = 2.34\,kPa- 100 \,kPa = -97.66 \, kPa\\

We insert that into our last equation and get:

\frac{ \Delta P}{ -g\, \rho\,}= \Delta h\\\Delta h=\frac{97.66 \, kPa}{998 kg/m^3 \, \, 9.81 m/s^2} \\\Delta h=9.975\,m

And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.

8 0
3 years ago
A(n)<br> is a safety device commonly<br> used with a slotted nut.
liraira [26]

A safety device called a cotter pin. The cotter pin fits through a hole in the bolt or part. This keeps the nut from turning and possibly coming off.

5 0
3 years ago
An inductor has a 50.0-Ω reactance when connected to a 60.0-Hz source. The inductor is removed and then connected to a 45.0-Hz s
nignag [31]

Given:

X_{L} = 50.0 \ohm

frequency, f = 60.0 Hz

frequency, f' = 45.0 Hz

V_rms} = 85.0 V

Solution:

To calculate max current in inductor, I_{L(max):

At f = 60.0 Hz

X_{L} = 2\pi fL

50.0 = 2\pi\times 60.0\times L

L = 0.1326 H

Now, reactance X_{L} at f' = 45.0 Hz:

X'_{L} = 2\pi f'L

X'_{L} = 2\pi\times 45.0\times 0.13263 = 37.5\ohm

Now, I_{L(max) is given by:

I_{L(max) = \sqrt {\frac{2V_{rms}}{X'_{L}}}

I_{L(max) = \sqrt {\frac{2\times 85.0}{37.5}} = 2.13 A

Therefore,  max current in the inductor, I_{L(max) = 2.13 A

7 0
3 years ago
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