<h3><u>CSMA/CD Protocol:
</u></h3>
Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.
But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.
There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.
<u>Example:
</u>
, at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.
12:00 AM A will see collisions
Pocket Size to detect the collision.
![\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26T_%7Bt%7D%20%5Cgeq%202%20T_%7BP%7D%5C%5C%26%5Cfrac%7BL%7D%7BB%7D%20%5Cgeq%202%20T_%7BP%7D%5C%5C%26L%20%5Cgeq%202%20%5Ctimes%20T_%7BP%7D%20%5Ctimes%20B%5Cend%7Baligned%7D)
CSMA/CD is widely used in Ethernet.
<u>Efficiency of CSMA/CD:</u>
- In the previous example we have seen that in worst case
time require to detect a collision.
- There could be many collisions may happen before a successful completion of transmission of a packet.
We are given number of collisions (contentions slots)=4.
Distance = 1km = 1000m
![\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%5Ctext%20%7B%20Speed%20%7D%3D2%20%5Ctimes%2010%5E%7B8%7D%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bsec%7D%5C%5C%20%26T_%7BP%7D%3D%5Cfrac%7B1000%7D%7B2%20%5Ctimes%2010%5E%7B8%7D%7D%3D%280.5%29%20%5Ctimes%2010%5E%7B-5%7D%3D5%20%5Ctimes%2010%5E%7B-6%7D%5C%5C%20%26T_%7Bt%7D%3D5%20%5Cmu%20%5Cmathrm%7Bsec%7D%5Cend%7Baligned%7D)
Correct answer is option E. No dimensions
As we know formula Pressure (P) is
also,
- Dimensional formula of <em>Pressure is </em>
![M^{1}L^{-1}T^{-2}](https://tex.z-dn.net/?f=M%5E%7B1%7DL%5E%7B-1%7DT%5E%7B-2%7D)
- Dimensional formula of <em>length is L </em>
- Dimensional formula of <em>mass is M</em>
- Dimensional formula of <em>velocity is </em>
![L^{1} T^{-1}](https://tex.z-dn.net/?f=L%5E%7B1%7D%20T%5E%7B-1%7D)
So, as given W=![\frac{P*L^{3} }{M*V^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7BP%2AL%5E%7B3%7D%20%7D%7BM%2AV%5E%7B2%7D%20%7D)
Dimensional formula of W =![\frac{M^{1}L^{-1}T^{-2} L^{3} }{M^{1} L^{2}T^{-2} }](https://tex.z-dn.net/?f=%5Cfrac%7BM%5E%7B1%7DL%5E%7B-1%7DT%5E%7B-2%7D%20%20L%5E%7B3%7D%20%20%7D%7BM%5E%7B1%7D%20L%5E%7B2%7DT%5E%7B-2%7D%20%20%20%7D)
since all terms get cancelled
Work is dimensionless i.e no dimensions
Learn more about dimensions here brainly.com/question/20351712
#SPJ10
Answer:
3.115×
meter
Explanation:
hall-petch constant for copper is given by
=25 MPa
k=0.12 for copper
now according to hall-petch equation
=
+![\frac{K}{\sqrt{D}}](https://tex.z-dn.net/?f=%5Cfrac%7BK%7D%7B%5Csqrt%7BD%7D%7D)
240=25+![\frac{0.12}{\sqrt{D}}](https://tex.z-dn.net/?f=%5Cfrac%7B0.12%7D%7B%5Csqrt%7BD%7D%7D)
D=3.115×
meter
so the grain diameter using the hall-petch equation=3.115×
meter
Answer:
Yes, fracture will occur
Explanation:
Half length of internal crack will be 4mm/2=2mm=0.002m
To find the dimensionless parameter, we use critical stress crack propagation equation
and making Y the subject
![Y=\frac {K}{\sigma_c \sqrt {a\pi}}](https://tex.z-dn.net/?f=Y%3D%5Cfrac%20%7BK%7D%7B%5Csigma_c%20%5Csqrt%20%7Ba%5Cpi%7D%7D)
Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness,
is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain
![Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682](https://tex.z-dn.net/?f=Y%3D%5Cfrac%20%7BK%7D%7B%5Csigma_c%20%5Csqrt%20%7Ba%5Cpi%7D%7D%3D%20%5Cfrac%20%7B40%7D%7B300%5Csqrt%20%7B%280.002%2A%5Cpi%29%7D%7D%3D1.682)
When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m
and making K the subject
and substituting 260 MPa for
while a is taken as 0.003m and Y is already known
![K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa](https://tex.z-dn.net/?f=K%3D260%2A1.682%2A%5Csqrt%20%7B0.003%2A%5Cpi%7D%3D42.455%20Mpa)
Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material
Answer:
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