Answer:
Absolute pressure , P(abs)= 433.31 KPa
Explanation:
Given that
Gauge pressure P(gauge)= 50 psi
We know that barometer reads atmospheric pressure
Atmospheric pressure P(atm) = 29.1 inches of Hg
We know that
1 psi = 6.89 KPa
So 50 psi = 6.89 x 50 KPa
P(gauge)= 50 psi =344.72 KPa
We know that
1 inch = 0.0254 m
29.1 inches = 0.739 m
Atmospheric pressure P(atm) = 0.739 m of Hg
We know that density of Hg =
P = ρ g h
P(atm) = 13.6 x 1000 x 9.81 x 0.739 Pa
P(atm) = 13.6 x 9.81 x 0.739 KPa
P(atm) =98.54 KPa
Now
Absolute pressure = Gauge pressure + Atmospheric pressure
P(abs)=P(gauge) + P(atm)
P(abs)= 344.72 KPa + 98.54 KPa
P(abs)= 433.31 KPa
Answer:
(a) water height =408.66 in.
(b) mercury height=30.04 in.
Explanation:
Given: P=14.769 psi ( 1 psi= 6894.76 
)
we know that 
where 
h=height.
Given that P=14.769 psi ⇒P= 101828.6 7
(a) 
⇒101828.67=
=10.38 m
So water barometer will read 408.66 in. (1 m=39.37 in)
(b) 
=13600
So 101828.67=
=0.763 m
So mercury barometer will read 30.04 in.
Answer:
The answer is c and the teacher helped me
Explanation:
i had help from the tescger and the assignment is done
<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
Answer:
0.304 L of Freon is needed
Explanation:
Q = mCT
Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J
C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K
T is temperature in the area of Mars = 189 K
m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg
Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3
Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L
