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expeople1 [14]
3 years ago
11

Technician A says the final drive assembly always has a gear ratio of 1:1. Technician B says the final drive assembly provides f

urther gear reduction between the transmission and the drive wheels. Who is correct?
Engineering
1 answer:
Olenka [21]3 years ago
5 0

Answer:

Technician B only is correct

Explanation:

The last stage of gears found between the vehicle transmission system and the wheels is the final drive ratio. The function of the final drive gear assembly is to enable a gear reduction control stage to reduce the rotation per minute and increase the wheel torque, such that the vehicle performance can be adjusted and the final gear ratio can be between 3:1 and 4.5:1 not 1:1

Therefore, technician B only is correct

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Give me the description of - Feedforward control loops with an example.
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Answer:

Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop

Explanation:

Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop. Although Feedforward control seems to be a very attractive idea, it imposes a high responsibility on both the system developer and the operator to examine and consider mathematically the effect of disruptions on the process concerned.

example of feedforward is  

Shower

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Customize water to compensate  

feedback refers to that point when water turns hot before the configuration changes

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Compute the number of games using the formula of “SINGLE ROUND ROBIN” and plot the games in a table:
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Answer:

N=27.02

Explanation:

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In 1945, the United States tested the world’s first atomic bomb in what was called the Trinity test. Following the test, images
Zarrin [17]

Answer:

r=K A^{1/5} \rho^{-1/5} t^{2/5}

A= \frac{r^5 \rho}{t^2}

A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT

Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

a) The energy A is released in a small place

b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant \rho_{air}.

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

[A]= \frac{ML^2}{T^2} A represent the energy and is defined as the mass times the velocity square, and the velocity is defined as \frac{L}{T}

[t]=T represent the time

[\rho]=\frac{M}{L^3} represent the density.

Solution to the problem 

And if we analyze the function for r we got this:

[r]=L=[A]^x [\rho]^y [t]^z

And if we replpace the formulas for each on we got:

[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z

And using algebra properties we can express this like that:

[r]=L=M^{x+y} L^{2x-3y} T^{-2x+z}

And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

x+y =0 , 2x-3y=1, -2x+z=0

We can solve for x like this x=-y and replacing into quation 2 we got:

2(-y)-3y = 1

-5y = 1

y= -\frac{1}{5}

And then we can solve for x and we got:

x = -y = -(-\frac{1}{5})=\frac{1}{5}

And if we solve for z we got:

z=2x =2 \frac{1}{5}=\frac{2}{5}

And now we can express the radius in terms of the dimensional analysis like this:

r=K A^{1/5} \rho^{-1/5} t^{2/5}

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}

A= \frac{r^5 \rho}{t^2}

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed \rho =1.2 kg/m^2, and replacing we got:

A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}

And we can convert this into ergs we got:

A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs

And then we know that 1 g of TNT have 4x10^4 erg

And we got:

A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT

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3 years ago
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Answer:

a. 121 Btu/lb

b. 211.8lb

c. 2.69/pc

Explanation:

See the attachments please

4 0
3 years ago
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