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Vaselesa [24]
3 years ago
9

before adjusting drive-belt tension, technician a checks for proper pulley alignment. technician b looks up the specified belt t

ension in the vehicle's service manual. who is correct?
Engineering
1 answer:
Vsevolod [243]3 years ago
5 0

Answer:

Technician b is correct

Explanation:

Before adjusting drive-belt tension, it is very important to check the vehicle workshop manual for specified belt tension, so that you can match your reading against the specification in the vehicle's service manual. If the tension reading you have matches the suggested reading in the vehicle's service manual and the belt is not damaged then you do not need to proceed any further. But if the reading does not match, then you can adjust the belt tension.

Therefore, technician b is correct.

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A very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one v
In-s [12.5K]

Answer:

The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s

Explanation:

Assuming the two oils are Newtonian fluids.

From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.

τ = μ (∂v/∂y)

There are oils above and below the plate, so we can write this expression for the both cases.

τ₁ = μ₁ (∂v/∂y)

τ₂ = μ₂ (∂v/∂y)

dv = 0.3 m/s

dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)

τ₁ = μ₁ (0.3/0.03) = 10μ₁

τ₂ = μ₂ (0.3/0.03) = 10μ₂

But the shear stress on the plate is given as 29 N per square meter.

τ = 29 N/m²

But this stress is a sum of stress due to both shear stress above and below the plate

τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29

But it is also given that one viscosity is twice the other

μ₁ = 2μ₂

10μ₁ + 10μ₂ = 29

10(2μ₂) + 10μ₂ = 29

30μ₂ = 29

μ₂ = (29/30) = 0.967 Pa.s

μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s

Hope this Helps!!!

6 0
3 years ago
This elementary problem begins to explore propagation delayand transmission delay, two central concepts in data networking. Cons
telo118 [61]

Explanation:

(a)

Here, distance between hosts A and B is m meters and, propagation speed along the link is s meter/sec

Hence, propagation delay, d_{prop} = m/sec (s)

(b)

Here, size of the packet is L bits

And the transmission rate of the link is R bps

Hence, the transmission time of the packet,  d_{trans} = L/R

(c)

As we know, end-to-end delay or total no delay,

\mathrm{d}_{\text {nodal }}=\mathrm{d}_{\text {proc }}+\mathrm{d}_{\text {quar }}+d_{\max }+d_{\text {prop }}

Here,  $\mathrm{d}_{\text {rroc }}$ and $\mathrm{d}_{\text {quat }}$ \\Hence, $\mathrm{d}_{\text {rodal }}=\mathrm{d}_{\text {trass }}+\mathrm{d}_{\text {prop }}$ \\We know, $\mathrm{d}_{\text {trax }}=\mathrm{L} / \mathrm{R}$ sec and $\mathrm{d}_{\text {vapp }}=\mathrm{m} / \mathrm{s}$ sec\text { Hence, } {d_{\text {nodal }}}=\mathrm{L} / \mathrm{R}+\mathrm{m} / \mathrm{s} \text { seconds }

(d)

The expression, time time $t=d_{\text {trans }}$ means the\at time since transmission started is equal to transmission delay.

As we know, transmission delay is the time taken by host to push out the packet.

Hence, at time $t=d_{\text {trans }}$ the last bit of the packet has been pushed out or transmitted.

(e)

If \ d_{prop} >d_{trans}

Then, at time $t=d_{\text {trans }}$ the bit has been transmitted from host A, but to condition (1),  the first bit has not reached B.

(f)

If \ d_{prop}

Then, at time $t=d_{\text {trans }}$, the first bit has reached destination on B

Here,s=2.5 \times 10^{8} \mathrm{sec}

\begin{aligned}&\mathrm{L}=100 \mathrm{Bits} \text { and }\\&\mathrm{R}=28 \mathrm{kbps} \text { or } 28 \times 1000 \mathrm{bps}\end{aligned}

It's given that \ d_{prop} =d_{trans}

Hence,

        \begin{aligned}\ & \frac{L}{R}=\frac{m}{s} \\m &=s \frac{L}{R} \\&=\frac{2.5 \times 10^{8} \times 100}{28 \times 1000} \\&=892.9 \mathrm{km}\end{aligned}

5 0
3 years ago
What are materials engineers trying to discover when they study different materials?
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What do materials science and engineering graduates do?
Almost two-thirds of materials science graduates are in employment six months after graduation.

The skills developed during a materials science degree allow graduates enter a range of sectors, including working as engineering professionals and in design and marketing roles.

Destination Percentage
Employed 60.4
Further study 24.5
Working and studying 5.1
Unemployed 4
Other 6.1
Graduate destinations for materials science and engineering
Type of work Percentage
Engineering and building 23.9
Marketing, PR and sales 11.9
Business, HR and financial 11.7
Technicians and other professionals 10
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