Question 6

1 answer:

5 0

A British thermal unit (BTU) is defined as the amount of heat required to raise the temperature of 1 pound of water by 1 degree Fahrenheit. To calculate the temperature of a water sample from the BTUs applied to it, you must know the weight of the water and its starting temperature. You can measure the weight of the water using a scale, and the temperature using a Fahrenheit thermometer. Once you have that information, calculating the temperature of the water after applying a known number of BTUs of heat is easy.

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The Ethernet (CSMA/CD) alternates between contention intervals and successful transmissions. Assume a 100 Mbps Ethernet over 1 k

Vesnalui [34]

<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>

$T_{P}=1 H r$ , 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.

$\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}$

CSMA/CD is widely used in Ethernet.

<u>Efficiency of CSMA/CD:</u>

In the previous example we have seen that in worst case $2 T_{P}$ 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.

$\text { Propagation day }=\frac{\text {distance}}{\text {speed}}$

Distance = 1km = 1000m

$\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}$

7 0

3 years ago

It is not a practical proposition to take direct measurements in nanoscale, but we can estimate variations in position and momen

Volgvan

Answer:

Answer is c Heisenberg's uncertainty principle

Explanation:

According to Heisenberg's uncertainty principle there is always an inherent uncertainty in measuring the position and momentum of a particle simultaneously.

Mathematically

$\Delta x\times \Delta \overrightarrow{p}\geq \frac{h}{4\pi }$