Which phrase best describes a safety-critical system? A. a system that faces a very high risk of failure B. a system isolated fr
1 answer:
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Answer:
percentage change in volume = 0.00285 %
Explanation:
given data
bulk modulus = 3.5 × N/m²
bulk stress = N/m²
solution
we will apply here bulk modulus formula that is
bulk modulus = ...............1
put here value and we get
3.5 × =
solve it we get
bulk strain = 2.8571 ×
and
bulk strain =
so that percentage change in volume is = 2.8571 × × 100
percentage change in volume = 0.00285 %
Answer:
a. 9947 m
b. 99476 times
c. 2*10^11 molecules
Explanation:
a) To find the mean free path of the air molecules you use the following formula:
R: ideal gas constant = 8.3144 Pam^3/mol K
P: pressure = 1.5*10^{-6} Pa
T: temperature = 300K
N_A: Avogadros' constant = 2.022*10^{23}molecules/mol
d: diameter of the particle = 0.25nm=0.25*10^-9m
By replacing all these values you obtain:
b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:
c) By using the equation of the ideal gases you obtain:
<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