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Anvisha [2.4K]
3 years ago
11

While having a discussion, Technician A says that you should never install undersized tires on a vehicle. The vehicle will be lo

wer, and the speedometer will no longer be accurate. Technician B says that the increase in engine rpm for a given speed will result in a decrease in fuel economy. Who is correct
Engineering
1 answer:
gulaghasi [49]3 years ago
6 0

Answer:

Both technician A and technician B are correct.

Explanation: Vehicle manufacturers always specify the size of the tires required for a given vehiclefor optimal efficiency,this will ensure that the speedometer is accurate and the level of the vehicle is good enough to ensure the vehicle works efficiently.

It is also a known fact that an increase in a vehicle's rpm(revolution per minute) will eventually lead to increased fuel consumption which means the fuel economy of the vehicle will be reduced making the vehicle less efficient in its fuel consumption.

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To put out a class D metal fire, you must _______ the fire.
gladu [14]

To put out a class D metal fire, you must smother the fire and eliminate the oxygen element in the fire.

<h3>What is a Class D fire?</h3>

A class D fire is a type of fire that cannot be extinguished by water. This is because adding water to it reacts with other elements in the fire intensifying the fire even more.

Smothering in this context involves adding a solution like carbon dioxide (CO2) into the fire, this results in a reduction of oxygen in the atmosphere surrounding the class D fire.

By so doing, smothering the fire eliminates the oxygen element in the fire, thereby extinguishing the fire.

You can learn more about extinguishing fires here https://brainly.in/question/760550

#SPJ1

7 0
2 years ago
A particle moving on a straight line has acceleration a = 5-3t, and its velocity is 7 at time t = 2. If s(t) is the distance fro
Vikki [24]

Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7

<h3>How to solve for the value of s2 - s1</h3>

We have

= \frac{dv}{dt} =v't = 5-3t\\\\\int\limits^a_b {v'(t)} \, dt

= \int\limits^a_b {(5-3t)} \, dt

5t - \frac{3t^2}{2} +c

v2 = 5x2 -  3x2 + c

= 10-6+c

= 4+c

s(t) = \frac{5t^2}{2} -\frac{t^3}{2} +3t + c

S2 - S1

=(5*\frac{4}{2} -\frac{8}{2} +3*2*c)-(\frac{5}{2} *1^2-\frac{1^2}{2} +3*1*c)

= 6 + 6+c - 2+3+c

12+c-5+c = 0

7 = c

Read more on acceleration here: brainly.com/question/605631

5 0
2 years ago
Explain why the following scenario fails to meet the criteria for proper reverse engineering.
avanturin [10]

Answer:

he must document or remember the order he took it apart so he put it back together

Explanation:

5 0
2 years ago
4. Water vapor enters a turbine operating at steady state at 1000oF, 220 lbf/in2 , with a volumetric flow rate of 25 ft3/s, and
hodyreva [135]
Yes i is the time of the day you get to frost the moon and back and then you can come over and then go to hang out with me me and then go to hang out
6 0
3 years ago
For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea
miskamm [114]

Answer:

k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

where r_1 and r_2 be the inner radius, outer radius of the annalus.

Explanation:

Let r_1, r_2 and L be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, T_1=30^{\circ}C\\ and at the outer surface, T_2=40^{\circ}C.

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

40 W/m^2.

\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40

\Rightarrow q=2.4\pi L \;W

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

q=-kA\frac{dT}{dr}\;\cdots(i)

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, A=2\pi rL\;\cdots(ii)

Variation of temperature w.r.t the radius of the annalus is

\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}

\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)

Putting the values from the equations (ii) and (iii) in the equation (i), we have

q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}

\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}

\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)} [as q=2.4\pi L, and T_2-T_1=10 ^{\circ}C]

\Rightarrow k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

7 0
3 years ago
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