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sladkih [1.3K]
3 years ago
6

Which of the following is the LEAST-Likely cause of tire wear? A) Underinflation B) Braking C) Acceleration D) Tire rotation

Engineering
1 answer:
mihalych1998 [28]3 years ago
4 0

Answer:

Tire rotation is the least likely cause of tire wear. So, the option D is correct.

Explanation:

Step1

Under-inflation is the process of tire failure under low pressure. This contributes the wear on tire.

Step2

On breaking, kinetic energy changes to heat energy because of rubbing of tire. So, rubbing action increases the wear on the tire.

Step3

Acceleration on the vehicle increases the rubbing action as well as the wear and tear on the tire. So, acceleration is an also a major cause of tire wear.

Step4

Tire rotation has least amount of wear and tear due to no rubbing action.  It has less amount surface contact with the surface in rotation.  

Thus, tire rotation is the least likely cause of tire wear. So, the option D is correct.  

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A liquid-liquid extraction process consists of two units, a mixer and a separator. One inlet stream to the mixer consists of two
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Answer:

One inlet stream to the mixer flows at 100.0 kg/hr and is 35wt% species-A and 65wt% species-B.

Explanation:

8 0
3 years ago
Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the
Musya8 [376]

Solution :

Given

$t_1=2+x_1$

$t_2=1+x_2$

Now,

$P(h

$0.4=1-P(h \geq5)$

$0.6=P(h \geq5)$

$0.6= e^{\frac{-x_1 5}{3600}}$

Therefore,   $x_1=368 \ veh/h$

                        $=\frac{368}{1000} = 0.368$

Given,   $t_1=2+x_1$

                 = 2 + 0.368

                 = 2.368 min

At user equilibrium, $t_2=t_1$

∴  $t_2$ = 2.368 min

$t_2=1+x_2$

$2.368=1+x_2$

$x_2 = 1.368$

$x_2 = 1.368 \times 1000$

    = 1368 veh/h

7 0
3 years ago
The advantages of solar cells include all of the following, except a.moderate net energy yield b.little or no direct emissions o
Xelga [282]

Answer:

C

Explanation:

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6 0
3 years ago
A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco
allsm [11]

Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

d_1=50 mm,f_1=0.0048

For pipe 2

d_2=75 mm,f_2=0.0058

Q=2.8 l/s

Q=2.8\times 10^{-3]

We know that Q=AV

Q=A_1V_1=A_2V_2

A_1=1.95\times 10^{-3}m^2

A_2=4.38\times 10^{-3} m^2

So V_2=0.63 m/s,V_1=1.43 m/s

head loss (h)

h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}

Now putting the all values

h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0
3 years ago
A fully braced structural member in a building is subjected to several different loads, including roof loads of D = 5 k and L_r
allochka39001 [22]

Answer: 37.4K

Explanation:

See attachment

7 0
3 years ago
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