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attashe74 [19]
3 years ago
6

Suppose you have an input gear that always turns clockwise. Suppose you have an output gear that does not engage the input gear

directly. Show, using diagrams, how you can design a gears train to generate:
a. A "Forward movement at the output.
b. A "Reverse" movement at the output.
Engineering
1 answer:
Norma-Jean [14]3 years ago
4 0

Answer:

Please find attached the required diagrams for;

a. A "Forward" movement at the output

b. A "Reverse" movement at the output

Explanation:

The direction of turning of the input gear = Clockwise

The engagement between the input and the output gear = Indirect engagement

a. For a "Forward" movement at the output, we have;

The number of idler gears between the input and output gear = 1

The direction of turning of the input gear = The direction the output gear turns

b. A "Reverse" movement at the output

For a "Reverse movement at the output, where the input and output gears do not directly engage. the number idler is increased is increased

∴ The number of idler between the input and output gears = 2

The direction of turning of the input gear = Opposite the direction the output gear turns

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Technician A says that the most effective way to diagnose wheel speed sensor circuit faults is to utilize the live data function
Marianna [84]

Answer:

Both Technicians A & B are correct

Explanation:

In diagnosing wheel systems, both steps said by technicians A & B are correct because it's the normal procedure to diagnose wheel speed sensor circuit faults.

7 0
3 years ago
The Environmental Protection Agency (EPA) has standards and regulations that says that the lead level in soil cannot exceed the
DENIUS [597]

Answer:

See below

Explanation:

<u>Check One-Sample T-Interval Conditions</u>

Random Sample? √

Sample Size ≥30? √

Independent? √

Population Standard Deviation Unknown? √

<u>One-Sample T-Interval Information</u>

  • Formula --> CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})
  • Sample Mean --> \bar{x}=390.25
  • Critical Value --> t^*=2.0096 (given df=n-1=50-1=49 degrees of freedom at a 95% confidence level)
  • Sample Size --> n=50
  • Sample Standard Deviation --> S_x=30.5

<u>Problem 1</u>

The critical t-value, as mentioned previously, would be t^*=2.0096, making the 95% confidence interval equal to CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})=390.25\pm2.0096(\frac{30.5}{\sqrt{50}})\approx\{381.5819,398.9181\}

This interval suggests that we are 95% confident that the true mean levels of lead in soil are between 381.5819 and 398.9181 parts per million (ppm), which satisfies the EPA's regulated maximum of 400 ppm.

3 0
2 years ago
Three spheres are subjected to a hydraulic stress. The pressure on spheres 1 and 2 is the same, and they are made of the same ma
r-ruslan [8.4K]

Answer:

"150000 N/m²" is the right approach.

Explanation:

According to the question, the pressure on the two spheres 1 and 2 is same.

Sphere 1 and 2:

Then,

⇒  P_1=P_2

⇒  \frac{\Delta V_1}{V_1}=\frac{\Delta V_2}{V_2}

and the bulk modulus be,

⇒  B_1=B_2

Sphere 3:

⇒  \frac{\Delta V_3}{V_3} =\frac{\frac{\Delta V_1}{V_1} }{\frac{\Delta V_2}{V_2} } =1

then,

⇒  P_3=B\times \frac{\Delta V_3}{V_3}

⇒       =B\times 1

⇒       =150000\times 1

⇒       =150000 \ N/m^2

5 0
3 years ago
A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the centr
seropon [69]

Answer:

1) The normal reactions at the front wheel is 9909.375 N

The normal reactions at the rear wheel is 8090.625 N

2) The least coefficient of friction required between the tyres and the road is 0.625

Explanation:

1) The parameters given are as follows;

Speed, u = 90 km/h = 25 m/s

Distance, s it takes to come to rest = 50 m

Mass, m = 1.8 tonnes = 1,800 kg

From the equation of motion, we have;

v² - u² = 2·a·s

Where:

v = Final velocity = 0 m/s

a = acceleration

∴ 0² - 25² = 2 × a × 50

a = -6.25 m/s²

Force, F =  mass, m × a = 1,800 × (-6.25) = -11,250 N

The coefficient of friction, μ, is given as follows;

\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625

Weight transfer is given as follows;

W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N

Therefore, we have for the car at rest;

Taking moment about the Center of Gravity CG;

F_R × 1.3 = 1.7 × F_F

F_R + F_F = 18000

F_R + \dfrac{1.3 }{1.7} \times  F_R = 18000

F_R = 18000*17/30 = 10200 N

F_F = 18000 N - 10200 N = 7800 N

Hence with the weight transfer, we have;

The normal reactions at the rear wheel F_R  = 10200 N - 2109.375 N = 8090.625 N

The normal reactions at the front wheel F_F =  7800 N + 2109.375 N = 9909.375 N

2) The least coefficient of friction, μ, is given as follows;

\mu = \dfrac{F}{R} =  \dfrac{11250}{18000} = 0.625

The least coefficient of friction, μ = 0.625.

3 0
4 years ago
Which option identifies the requirement standard NOT met in the following scenario?
LenaWriter [7]

Answer:

The requirement does not meet the feasibility standard.

Explanation:

because the people can just put on the blackbelts

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3 years ago
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