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4 years ago

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If datums are used as suggested below (A datum on shaft, B datum on face below the head, C datum on the hex flat) and order of p

recedence is in alphabetic order, name the datum that prevents: a. Translation along the X axis: _ b. Rotation about the X axis: _ c. Translation along the Z axis: _ d. Rotation about the Y axis: _

1 answer:

allsm [11]4 years ago

6 0

Answer:

a. Use datum on shaft

b. Use datum on hex flat

c. Use datum on face below the head

d. Use datum on shaft

When these datum are used, they will prevent translation and rotation along axis which they act.

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This elementary problem begins to explore propagation delayand transmission delay, two central concepts in data networking. Cons

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Explanation:

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

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

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Hence,

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

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3 years ago

What are the basic steps in the operation of a solar cell?

kati45 [8]

Answer:

Light is absorbed and knocks electrons loose. Loose electrons flow, creating a current. The current is captured and transferred to wires.

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3 years ago

A well-insulated rigid tank contains 5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters of

vlabodo [156]

Answer:

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$S_{sys,1} - S_{sys,2} + S_{gen} = 0$

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$S_{gen} = m\cdot (s_{2}-s_{1})$

The properties of the steam are obtained from steam tables:

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$P = 200\,kPa$

$T = 120.21\,^{\textdegree}C$

$\nu = 0.2222\,\frac{m^{3}}{kg}$

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Answer:

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Consider the following two well-mixed, isothermal gas-phase batch reactors for the elementary and irreversible decomposition of

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Answer:

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