The weight of the specimen in SSD condition is 373.3 cc
<u>Explanation</u>:
a) Apparent specific gravity = ![\frac{A}{A-C}](https://tex.z-dn.net/?f=%5Cfrac%7BA%7D%7BA-C%7D)
Where,
A = mass of oven dried test sample in air = 1034 g
B = saturated surface test sample in air = 1048.9 g
C = apparent mass of saturated test sample in water = 975.6 g
apparent specific gravity =
= ![\frac{1034}{1034-675 \cdot 6}](https://tex.z-dn.net/?f=%5Cfrac%7B1034%7D%7B1034-675%20%5Ccdot%206%7D)
Apparent specific gravity = 2.88
b) Bulk specific gravity ![G_{B}^{O D}=\frac{A}{B-C}](https://tex.z-dn.net/?f=G_%7BB%7D%5E%7BO%20D%7D%3D%5Cfrac%7BA%7D%7BB-C%7D)
![G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}](https://tex.z-dn.net/?f=G_%7BB%7D%5E%7BO%20D%7D%3D%5Cfrac%7B1034%7D%7B1048.9-675%20%5Ccdot%206%7D)
= 2.76
c) Bulk specific gravity (SSD):
![G_{B}^{S S D}=\frac{B}{B-C}](https://tex.z-dn.net/?f=G_%7BB%7D%5E%7BS%20S%20D%7D%3D%5Cfrac%7BB%7D%7BB-C%7D)
![=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1048%20%5Ccdot%209%7D%7B1048%20%5Ccdot%209-675%20%5Ccdot%206%7D)
= 2.80
d) Absorption% :
![=\frac{B-A}{A} \times 100 \%](https://tex.z-dn.net/?f=%3D%5Cfrac%7BB-A%7D%7BA%7D%20%5Ctimes%20100%20%5C%25)
![=\frac{1048 \cdot 9-1034}{1034} \times 100](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1048%20%5Ccdot%209-1034%7D%7B1034%7D%20%5Ctimes%20100)
Absorption = 1.44 %
e) Bulk Volume :
![v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}](https://tex.z-dn.net/?f=v_%7Bb%7D%3D%5Cfrac%7B%5Ctext%20%7B%20weight%20of%20dispaced%20water%20%7D%7D%7BP%20%5Comega%20t%7D)
![=\frac{1048 \cdot 9-675 \cdot 6}{1}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1048%20%5Ccdot%209-675%20%5Ccdot%206%7D%7B1%7D)
= ![373.3 cc](https://tex.z-dn.net/?f=373.3%20cc)
Answer:
a) ![\frac{Ws}{Es} = \frac{200}{1+1.2s}](https://tex.z-dn.net/?f=%5Cfrac%7BWs%7D%7BEs%7D%20%20%3D%20%5Cfrac%7B200%7D%7B1%2B1.2s%7D)
b) attached below
c) type zero system
d) k > ![\frac{g}{200}](https://tex.z-dn.net/?f=%5Cfrac%7Bg%7D%7B200%7D)
e) The gain K increases above % error as the steady state speed increases
Explanation:
Given data:
Motor voltage = 12 v
steady state speed = 200 rad/s
time taken to reach 63.2% = 1.2 seconds
<u>a) The transfer function of the motor from voltage to speed</u>
let ;
be the transfer function of a motor
when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec
hence the transfer function of the motor from voltage to speed
= ![\frac{Ws}{Es} = \frac{200}{1+1.2s}](https://tex.z-dn.net/?f=%5Cfrac%7BWs%7D%7BEs%7D%20%20%3D%20%5Cfrac%7B200%7D%7B1%2B1.2s%7D)
<u>b) draw the block diagram of the system with plant controller and the feedback path </u>
attached below is the remaining part of the detailed solution
c) The system is a type-zero system because the pole at the origin is zero
d) ) k > ![\frac{g}{200}](https://tex.z-dn.net/?f=%5Cfrac%7Bg%7D%7B200%7D)
Answer: the half-angle "alpha" of the Mach cone = 30⁰
Explanation:
To calculate the half-angle "alpha" of the Mach cone.
we say ;
Sin∝ = 1 / Ma
given that Ma = 2
now we substitute
Sin∝ = 1 / 2
Sin∝ = 0.5
∝ = Sin⁻¹ 0.5
∝ = 30⁰
Therefore, the half-angle "alpha" of the Mach cone is 30⁰