Answer:
938.7 milliseconds
Explanation:
Since the transmission rate is in bits, we will need to convert the packet size to Bits.
1 bytes = 8 bits
1 MiB = 2^20 bytes = 8 × 2^20 bits
5 MiB = 5 × 8 × 2^20 bits.
The formula for queueing delay of <em>n-th</em> packet is : (n - 1) × L/R
where L : packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate = 2.1 Gbps = 2.1 × 10^9 bits per second.
Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9
queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9
queueing delay for 48th packet = 0.938725181 seconds
queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds
5 is the correct one to choose for this
Answer:
maximum isolator stiffness k =1764 kN-m
Explanation:
mean speed of rotation 
=65.44 rad/sec
= 0.1*(65.44)^2
F_T =428.36 N
Transmission ratio 
also
transmission ratio ![= \frac{1}{[\frac{w}{w_n}]^{2} -1}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B%5B%5Cfrac%7Bw%7D%7Bw_n%7D%5D%5E%7B2%7D%20-1%7D)
![0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}](https://tex.z-dn.net/?f=0.7%20%3D%5Cfrac%7B1%7D%7B%5B%5Cfrac%7B65.44%7D%7Bw_n%7D%5D%5E2%20-1%7D)
SOLVING FOR Wn
Wn = 42 rad/sec
k = m*W^2_n
k = 1000*42^2 = 1764 kN-m
k =1764 kN-m
Answer:
I'm no engineer, but blue and purple are cool colors and white is every color so I'd go with orange
Answer:
401.3 kg/s
Explanation:
The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).
qw = 0.85 * q2
q2 = 0.64 * q1
p = 0.36 * q1
q1 = p /0.36
q2 = 0.64/0.36 * p
qw = 0.85 *0.64/0.36 * p
qw = 0.85 *0.64/0.36 * 600 = 907 MW
In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)
The latent heat for the vaporization of water is:
SLH = 2.26 MJ/kg
So, to dissipate 907 MW
G = qw * SLH = 907 / 2.26 = 401.3 kg/s
