Answer:
5.21e-2mm
Explanation:
Please see attachment
Suppose there are 93 packets entering a queue at the same time. Each packet is of size 4 MiB. The link transmission rate is 1.4
1 answer:
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Answer:
8861.75 m approximately 8862 m
Explanation:
We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.
for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that
For simplicity we work with
despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.
In accord with the formula the "normal" or "standard" weight of an object is given by
when
, so we need to find the value of
that makes
meaning that the original weight decrease by a 0.30%, so now we operate...
now we group like terms on the same sides
we cancel equal tems on both sides and obtain that
Answer:
a) Tբ = 151.8°C
b) ΔV = - 0.194 m³
c) The T-V diagram is sketched in the image attached.
Explanation:
Using steam tables,
At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.
Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C
b) The volume change
Using data from A-5 and A-6 of the steam tables,
The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume
(αբ) (which is calculated from the final quality and the consituents of the specific volumes).
ΔV = m(αբ - αᵢ)
αբ = αₗ + q(αₗᵥ) = αₗ + q (αᵥ - αₗ)
q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg
αբ = 0.00109 + 0.5(0.3748 - 0.00109)
αբ = 0.187945 m³/kg
αᵢ = 0.5226 m³/kg
ΔV = 0.58 (0.187945 - 0.5226) = - 0.194 m³
c) The T-V diagram is sketched in the image attached
