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3 years ago

5

Technician A says that rear-wheel drive vehicles usually get better traction than front-wheel drive vehicles. Technician B says

that front-wheel drive vehicles get extremely good traction due to the weight of the engine and transaxle. Who is correct?
a

Technician A

b

Technician B

c

Both A & B

d

Neither A or B

1 answer:

cluponka [151]3 years ago

3 0

C is the correct answer

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End A of the uniform 5-kg bar is pinned freely to the collar, which has an acceleration = 4 m/s2 along the fixed horizontal shaf

sergiy2304 [10]

Answer:

Fa = 57.32 N

Explanation:

given data

mass = 5 kg

acceleration = 4 m/s²

angular velocity ω = 2 rad/s

solution

first we take here moment about point A that is

∑Ma = Iα + ∑Mad ...............1

put here value and we get

so here I = ( $\frac{1}{12}$ ) × m × L² ................2

I = ( $\frac{1}{12}$ ) × 5 × 0.8²

I = 0.267 kg-m²

and

a is = r × α

a = 0.4 α

so now put here value in equation is 1

0 = 0.267 α + m r α (0.4) - m A (0.4)

0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4

so angular acceleration α = 7.5 rad/s²

so here force acting on x axis will be

∑ F(x) = m a(x) ..............3

a(x) = m a - m rα

put here value

a(x) = 5 × 4 - 5 × 0.4 × 7.5

a(x) = 5 N

and

force acting on y axis will be

∑ F(y) = m a(y) .............. 4

a(y) - mg = mrω²

a(y) - 5 × 9.81 = 5 × 0.4 × 2²

a(y) = 57.1 N

so

total force at A will be

Fa = $\sqrt{a(x)^2+a(y)^2}$ ...............5

Fa = $\sqrt{(57.1)^2+(5)^2}$

Fa = 57.32 N

3 0

4 years ago

If you think a hazard is serious, what is the best way to ensure that OSHA will

Umnica [9.8K]

Answer:

The best way to ensure that OSHA will conduct a site inspection is by the submission a signed (by a current employee or employee representative) Formal Written complaints following OSHA guidelines (which can be found in the Council of Safety and Health (COSH) website) which will enable the commencement of process for an on site inspection

As such it is advisable to go through the following procedure before, filing a complaint

1) Talk to your workers union about the hazard

2) If you do not belong to a union, talk to your co-workers

3) Inform your employer who has responsibility for on the job safety

4) Speak or have a meeting with OSHA so as to receive guidance before filing a complaint

Explanation:

3 0

3 years ago

Consider liquid n-hexane in a 50-mm diameter graduated cylinder. Air blows across the top of the cylinder. The distance from the

ra1l [238]

The evaporation rate of the n-Hexane is $7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}$

<u>Explanation</u>:

This is a situation regarding diffusing A through non-diffusing B.

A = n-Hexane B=Air

Where the molar flux is provided by,

$N_{A}=D_{A B} P_{T}\left(P_{A 1}-P_{A 2}\right) / R T z P_{b m}$

$\mathrm{D}_{\mathrm{AB}}=8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}$

$P_{t}=1 a t m=101325 P a\\$

$\text { so, } P_{A 1}=$ the vapor pressure at hexane $25 \mathrm{C}$ $=20158.2 \mathrm{Pa}$

For wind, assume negligible hexane is present, hence $P_{A 2}=0$

Now,

$\mathrm{P}_{\mathrm{B} 1}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 1}=101325-20158.2 \mathrm{P}_{\mathrm{a}}$

$\mathrm{P}_{\mathrm{B} 2}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 2}=\mathrm{P}_{\mathrm{T}}=101325 \mathrm{Pa}$

$P_{B M}=\frac{\left(P_{B 2}-P_{B 1}\right)}{\log _{e}\left(P_{B 2} / P_{B 1}\right)}\\$

$=\frac{101325-81166.8}{\ln \left(\frac{101325}{81166.8}\right) \mathrm{Pa}}$

$=90873.57 \mathrm{Pa}$

$R=8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K}$

$z=\text { distance }=20 \mathrm{cm}=0.2 \mathrm{m}\\$

where T = 298 K

substituting all in the equation, we get

$\begin{aligned}&\mathrm{N}_{\mathrm{A}}=\\&\left(8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\right) \times 101325 \mathrm{Pa} \times(20158.2 \mathrm{Pa}) /(8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K} \times 0.2 \mathrm{m} \times 298 \mathrm{K}\\&\times 90873.57 \mathrm{Pa})\end{aligned}$

$=0.004 \mathrm{mol} / \mathrm{m}^{2} \mathrm{s}\\$

Now,Flux $\times$ area = Molar rate of evaporation

Evaporation rate = $0.004 \mathrm{mol} / \mathrm{m}^{2}-5 \mathrm{x}\left(\pi \mathrm{d}^{2} / 4 \mathrm{m}^{2}\right)=0.004 \times(3.14 \times 0.05 \times 0.05 / 4)$

Evaporation rate = $7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}$

6 0

3 years ago

List three types of concurrent engineering in manufacturing.

klio [65]

Answer:

A famous example of concurrent engineering is the development of the Boeing 777 commercial aircraft. The aircraft was designed and built by geographically distributed companies that worked entirely on a common product database of C A TIA without building physical mock-ups but with digital product definitions.

8 0

3 years ago

This is just so I can flip the picture

Pepsi [2]

Answer:

ok

Explanation:

thx for points

7 0

4 years ago