All categories

3 years ago

How are the suspension systems related to the steering system?

Engineering

2 answers:

Novosadov [1.4K]3 years ago

7 0

Answer:
Yes
Step by step explication:

harina [27]3 years ago

3 0

Answer:

Rack and Pinion Steering Rack and pinion steering transmits circular motion from the steering wheel to a pinion that meshes with teeth on a flat rack. ... Suspension Systems The suspension system supports the vehicle, allowing the wheels to move up and down over irregularities in the road.

You might be interested in

Which of the following might be a job or task of an IT worker who manages

-Dominant- [34]

Answer:

I'm pretty sure it's letter d

7 0

3 years ago

In a planetary geartrain with a form factor of 8, the sun gear rotates clockwise at 5 rad⁄s and the ring gear rotates clockwise

lina2011 [118]

Answer:

D. N= 11. 22 rad/s (CW)

Explanation:

Given that

Form factor R = 8

Speed of sun gear = 5 rad/s (CW)

Speed of ring gear = 12 rad/s (CW)

Lets take speed of carrier gear is N

From Algebraic method ,the relationship between speed and form factor given as follows

$\dfrac{N_{sun}-N}{N_{ring}-N}=-R$

here negative sign means that ring and sun gear rotates in opposite direction

Lets take CW as positive and ACW as negative.

Now by putting the values

$\dfrac{N_{sun}-N}{N_{ring}-N}=-R$

$\dfrac{5-N}{12-N}=-8$

N= 11. 22 rad/s (CW)

So the speed of carrier gear is 11.22 rad/s clockwise.

8 0

3 years ago

For goods-producing firms, at which of the following levels of resource planning does scheduling for individual subassemblies an

VashaNatasha [74]

Answer:

Disaggregation

Explanation:

In a company it is a way to create operational plans that are focused, either by time or by section.

3 0

3 years ago

Whats 47,000 resistance converted to kilo or mega ohms.

Julli [10]

The conversion of 47,000 Ohms to kilo-ohms is equal to 47 kilo-ohms.
The conversion of 47,000 Ohms to mega-ohms is equal to 0.047 kilo-ohms.

<h3>What is resistance?</h3>

Resistance can be defined as an opposition to the flow of current in an electric circuit. Also, the standard unit of measurement of the resistance of an electric component is Ohms, which can be converted to kilo-ohms or mega-ohms.

For Ohms to kilo-ohms, we have:

1 Ohms = 0.001 kilo-ohms

47,000 Ohms = X kilo-ohms

Cross-multiplying, we have:

X = 0.001 × 47000

X = 47 kilo-ohms.

For Ohms to mega-ohms, we have:

1,000,000 ohms = 1 mega-ohms

47,000 Ohms = X mega-ohms

Cross-multiplying, we have:

X1,000,000 = 47,000

X = 47,000/1,000,000

X = 0.047 kilo-ohms.

Read more resistance here: brainly.com/question/19582164

#SPJ1

7 0

1 year ago

Flank wear data were collected in a series of turning tests using a coated carbide tool on hardened alloy steel at a feed of 0.3

Paladinen [302]

Answer:

A) n = 0.6143, c ≈ 640m/min

B) n = 0.6143 , c = 637.53m/min

Explanation:

using the given data

A) A plot of flank wear as a function of time and also A plot for tool when

Flank wear is 0.75 and cutting edge speed is 100m/min, Time of cutting edge is said to be 20.4 min also for cutting edge speed of 155m/min , time for cutting edge is 10 min

is attached below

calculate for the constant N from the second plot

note : the slope will be negative because cutting speed decreases as time of cutting increase

V1 = 100m/min , V2 = 155m/min, T1 = 20.4 min, T2 = 10 min

= - N = $\frac{In(V2) - In(V1)}{In(T2)-ln(T1)}$

therefore - N = $\frac{5.043 - 4.605}{2.302 -3.015}$

= - 0.6143

THEREFORE ( N ) = 0.6143

Determine for the constant C from the second plot as well

note : C is the intercept on the cutting speed axis in 1 min tool life

connecting the two points with a line and extend it to touch the cutting speed axis and measure the value at that point

hence C ≈ 640m/min

B) Calculate the values of N and C in the Taylor equation solving simultaneous equations

using the above cutting speed and time of cutting values we can find the constant N via Taylor tool life equation

Taylor tool life equation = vT = C ------------- equation 1

cutting speed = v = 100m/min and 155m/min

tool life = T = 20.4 min and 10 min

also constant n and c are obtained from the previous plot

back to taylor tool life equation = 100 * 20.4 = C

therefore C = (100)(20.4)^n ---------------- equation 2

also using the second values of v and T

taylor tool life equation = 155 * 10 = C

therefore C = ( 155 )(10)^n ----------------- equation 3

Equate equation 2 and equation 3 and solve simultaneously

(100)(20.4)^n = (155)(10)^n

To find N

take natural log of both sides of the equation

= In ((100)(20.4)^n) = In((155)(10)^n)

= In (100) + nIn(20.4) = In(155) + nIn(10)^n

= n(3.0155) - n (2.3026) = 5.043 - 4.605

= 0.7129 n = 0.438

therefore n = 0.6143

To find C

substitute 0.6143 for n in equation 2

C = (100)(20.4) ^ 0.6143

C = 637.53 m/min

Attached are the two plots for solution A

7 0

3 years ago