A particular DSL modem operates at 768 kbits/sec. How many bytes can it receive in 1 minute? USB 3.0 can send data at 5 Gbits/se
1 answer:
Answer:
a. 1.6 Kbytes/min
b. 10.417 Mbytes/min
Explanation:
a. The DSL modem operates at 768 Kbits/sec.
But,
8 bits = 1 byte and 1 Kbit = 1 000 bits, so that:
=
= 96 000 bytes
Therefore, the modem operates at 96 Kbytes/sec.
The byte to be received in 1 minute can be calculated thus;
Since 60 seconds = 1 minute, then:
=
= 1600
= 1.6 Kbytes/min
The modem receives 1.6 Kbytes/min
b. The USB sends 5 Gbits/sec.
But, 8 bits = 1 byte and 1 Gbit = 1000000000 bits so that:
=
= 625000000
= 0.625 Gbytes
The USB sends 0.625 Gbyte/sec.
Since 60 seconds = 1 minute, then:
=
= 10416666.67
= 10.417 Mbytes/min
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Answer:
the rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.
Explanation:
Given the data in the question;
From the first law of thermodynamics;
dQ = dU + dW ------ let this be equation 1
where dQ is the heat transfer, dU is internal energy and dW is the work done.
from the question, the process is isothermal ( internally reversible process )
Thus, the change in internal energy is 0
dU = 0
given that; Air is compressed by a 40-kW compressor from P1 to P2
since it is compressed, dW = -40 kW
we substitute into equation 1
dQ = 0 + ( -40 kW )
dQ = -40 kW
Now, change in entropy of air is;
ΔS = dQ / T
given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K
so we substitute
ΔS = -40 kW / 298.15 K
ΔS = -0.13416 ≈ -0.1342 kW/K
Therefore, the rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.