Answer:
I know it is C)Virtual reality
Explanation:
Look at the clues
story about putting on a headset ( virtual reality head set!)
seeing a digital world (A virtual reality world)
they could walk around in (Fake walking you are basically jogging in place)
explore in order to see what ancient Benin looked like (Looking at a real place only digitally)
as if they were really there ( they think they are actually there)
The only reason I know all of this is because I have done virtual reality multiple times and I LOVED it SUPER fun ( I was doing archery) :) Hope this helps!
Answer:
Power required to overcome aerodynamic drag is 50.971 KW
Explanation:
For explanation see the picture attached
Answer:
The power developed in HP is 2702.7hp
Explanation:
Given details.
P1 = 150 lbf/in^2,
T1 = 1400°R
P2 = 14.8 lbf/in^2,
T2 = 700°R
Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h
Using air table to obtain the values for h1 and h2 at T1 and T2
h1 at T1 = 1400°R = 342.9 Btu/h
h2 at T2 = 700°R = 167.6 Btu/h
Using;
Q - W + m(h1) - m(h2) = 0
W = Q - m (h2 -h1)
W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h
W = (-65000 Btu/h ) - (-1928.3) Btu/s
W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s
W = -18.06Btu/s + 1928.3 Btu/s
W = 1910.24Btu/s
Note; Btu/s = 1.4148532hp
W = 2702.7hp
Answer:
Football stadium on rocky soil
Skyscraper on bedrock
Apartment building on sandy soil
Explanation:
Answer:
a.)
US Sieve no. % finer (C₅ )
4 100
10 95.61
20 82.98
40 61.50
60 42.08
100 20.19
200 6.3
Pan 0
b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4
c.) Cu = 3.33
d.) Cc = 1
Explanation:
As given ,
US Sieve no. Mass of soil retained (C₂ )
4 0
10 18.5
20 53.2
40 90.5
60 81.8
100 92.2
200 58.5
Pan 26.5
Now,
Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g
⇒ w = 421.2 g
As we know that ,
% Retained = C₃ = C₂×![\frac{100}{w}](https://tex.z-dn.net/?f=%5Cfrac%7B100%7D%7Bw%7D)
∴ we get
US Sieve no. % retained (C₃ ) Cummulative % retained (C₄)
4 0 0
10 4.39 4.39
20 12.63 17.02
40 21.48 38.50
60 19.42 57.92
100 21.89 79.81
200 13.89 93.70
Pan 6.30 100
Now,
% finer = C₅ = 100 - C₄
∴ we get
US Sieve no. Cummulative % retained (C₄) % finer (C₅ )
4 0 100
10 4.39 95.61
20 17.02 82.98
40 38.50 61.50
60 57.92 42.08
100 79.81 20.19
200 93.70 6.3
Pan 100 0
The grain-size distribution is :
b.)
From the diagram , we can see that
D10 = 0.12
D30 = 0.22
D60 = 0.12
c.)
Uniformity Coefficient = Cu = ![\frac{D60}{D10}](https://tex.z-dn.net/?f=%5Cfrac%7BD60%7D%7BD10%7D)
⇒ Cu = ![\frac{0.4}{0.12} = 3.33](https://tex.z-dn.net/?f=%5Cfrac%7B0.4%7D%7B0.12%7D%20%3D%203.33)
d.)
Coefficient of Graduation = Cc = ![\frac{D30^{2}}{D10 . D60}](https://tex.z-dn.net/?f=%5Cfrac%7BD30%5E%7B2%7D%7D%7BD10%20.%20D60%7D)
⇒ Cc =
= 1