3 years ago

Excessive looseness in steering and suspension components can cause _____

Engineering

1 answer:

Phoenix [80]3 years ago

5 0

Answer:

dynamic imbalance may cause the steering wheel to pull both directions

Explanation:

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Joe has a set up in which the length of the load arm is 20cm and the length of the effort arm is 80cm. What is the mechanical ad

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Answer:

Explanation:

Mechanical advantage = effort arm / load arm

MA = 80 cm / 20 cm

MA = 4

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3 years ago

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You push a 10 n object 10 m how much work was done on the object

dezoksy [38]

Answer: the answer is

Work= Force * Distance

Work= 10N * 10m

Work= 100J

Explanation:

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3 years ago

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kow [346]

Answer:

i wish i knew what you meant by that

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3 years ago

Two soils are fully saturated with liquid (no gas present) and the soils have the same void ratio. One soil is saturated with wa

mariarad [96]

Answer:

water sample have more water content

Explanation:

given data

soil 1 is saturated with water

unit weight of water = 1 g/cm³

soil 2 is saturated with alcohol

unit weight of alcohol = 0.8 g/cm³

solution

we get here water content that is express as

water content = $S_s \times \gamma _w$ ....................1

here soil is full saturated so $S_s$ is 100% in both case

so put here value for water

water content = 100 % × 1

water content = 1 g

and

now we get for alcohol that is

water content = 100 % × 0.8

water content = 0.8 g

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4 years ago

Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distanc

Andrei [34K]

Answer:

the answer is below

Explanation:

a) The conductivity of graphite (σ) is calculated using the formula:

$\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}$

where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012

Substituting:

$\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m$

b) f = 1 GHz = 10⁹ Hz.

$\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10} e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10} e=-1.5 \\\\z=\frac{-1.5}{log_{10} e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm$

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3 years ago