Answer:
I am in 6th grade, why are high school things popping up??
Explanation:
Answer:
The distance between the centers of adjacent atoms for the FCC crystal structure along the [100] is 2R√2
Explanation:
From the image uploaded, a Face centered cubic structure (100) plane, there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells, while the center atom lies entirely within the unit cell.
In terms of the atomic radius, R, we determine the distance between the centers of adjacent atoms.
Let this distance = AC
the two adjacent sides = AB and BC
AB = a = 2R
BC = a = 2R
Using Pythagoras theorem
AC² = AB² + BC²
AC² = a² + a²
AC² = 2a²
AC = √2a²
AC = a√2
But a = 2R
AC = 2R√2
Therefore, the distance between the centers of adjacent atoms for the FCC crystal structure along the [100] is 2R√2
<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.

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

A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s) with a flow rate of 8.7 m^3/s, 6 mg/L BOD5 and 8.3 mg/L DO. Both streams are at 20°C. After mixing, the river is 3 meters deep and flowing at a velocity of 0.50 m/s. DOsat for this river is 9.0 mg/L. The deoxygenation constant is kd= 0.20 d^-1 and The reaction rate constant k at 20 °C is 0.27 d^-1.
The answer therefore would be the number 0.27 divided by two and then square while getting the square you would make it a binomial.
I wont give the answer but the steps
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