Answer:
True
Explanation:
Actually this are some of the nitty gritty answers and ways to control or stip stress, lemme explain them you see stress most of the time may come from deep thoughts that are and are pushing you to the wall, and you in your state you as you react to that which is so demanding as it makes you tense,so if at you will incorporate physical activities like running, jogging or push ups the psychological tension in you is broken as you focus so much on the activities rather than the tension, then getting enough rest cools the mind and all your thoughts settle as in your brain starts to adapt to chilling and relaxation and enough sleep will actually make ones head to be at peace because if you lack enough sleep you might have an excruciating migraine when you are stressed up and finally relaxation techniques makes the body accept the situation and then you manouver out of it as you grow strongly.
Hope this will help!
Answer:
The concentration in mg/L equals 0.01 mg/L
Explanation:
Since the concentration is given as 10 parts per billion
We know that
1 part per billion equals 0.001 mg/L
Thus 10 parts per billion concentration equals
Answer:
number of houses = 3751.243
Explanation:
given data
tower high H = 90 ft
pipe length L = 3 mile
pipe dia d = 6 in
solution
we consider here loss is neglected by dia 6 in pipe
so we apply here bernaulis equation from top to bottom height 90 ft
..........................1
here P1 is = o gauge pressure
and P2 = 30 Psi = 206.843 × Pa
and Z1 = 27.432 m
and Z2 = 0 and V1 = 0
so from equation 1
0+0+27.432 = ×
solve we get
V = 11.16 m/s
V = 36.6 ft/s
and
flow will be here
flow Q = AV ............2
Q = × 11.16
Q = 0.19723 m³/s
Q = 187562.157 gal/hr
we have given house use maximum = 50 gal/hr
so total home served =
number of houses =
so number of houses = 3751.243
Answer:
0.19s
Explanation:
Queueing delay is the time a job waits in a queue before it can be executed. it is the difference in time betwen when the packet data reaches it destination and the time when it was executed.
Queueing delay =(N-1) L /2R
where N = no of packet =93
L = size of packet = 4MB
R = bandwidth = 1.4Gbps = 1×10⁹ bps
4 MB = 4194304 Bytes
(93 - 1)4194304 / 2× 10⁹
queueing delay =192937984 ×10⁻⁹
=0.19s