Okay sure.
I’ll 1)chords
2)pulse
3)aerophone
4) the answer is C
5)rhythm
Pretty sure those are the answers
Answer:
The time required is 10.078 hours or 605 min
Explanation:
The formula to apply here is ;
K=(d²-d²₀ )/t
where t is time in hours
d is grain diameter to be achieved after heating in mm
d₀ is the grain diameter before heating in mm
Given
d=5.5 × 10^-2 mm
d₀=2.4 × 10^-2 mm
t₁= 500 min = 500/60 =25/3 hrs
t₂=?
n=2.2
First find K
K=(d²-d²₀ )/t₁
K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3
K=(0.051²-0.024²) ÷25/2
K=0.000243 mm²/h
Re-arrange equation for K ,to get the equation for d as;
d=√(d₀²+ Kt) where now t=t₂

Answer:
1) 
2) 
Explanation:
For isothermal process n =1

![V_o = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}](https://tex.z-dn.net/?f=V_o%20%20%3D%20%5Cfrac%7B5%7D%7B%5B%5Cfrac%7B72%7D%7B80%7D%5D%5E%7B1%2F1%7D%20-%5B%5Cfrac%7B72%7D%7B180%7D%5D%5E%7B1%2F1%7D%7D)

calculate pressure ratio to determine correction factor

correction factor for calculate dpressure ration for isothermal process is
c1 = 1.03

b) for adiabatic process
n =1.4
volume of hydraulic accumulator is given as
![V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}](https://tex.z-dn.net/?f=V_o%20%3D%5Cfrac%7B%5CDelta%20V%7D%7B%5B%5Cfrac%7Bp_o%7D%7Bp_1%7D%5D%5E%7B1%2Fn%7D%20-%5B%5Cfrac%7Bp_o%7D%7Bp_2%7D%5D%5E%7B1%2Fn%7D%7D)
![V_o = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}](https://tex.z-dn.net/?f=V_o%20%20%3D%20%5Cfrac%7B5%7D%7B%5B%5Cfrac%7B72%7D%7B80%7D%5D%5E%7B1%2F1.4%7D%20-%5B%5Cfrac%7B72%7D%7B180%7D%5D%5E%7B1%2F1.4%7D%7D)

calculate pressure ratio to determine correction factor

correction factor for calculate dpressure ration for isothermal process is
c1 = 1.15

Answer:
Explanation:
The detailed steps is as shown in the attachment.
Explanation:
Science is the body of knowledge that explores the physical and natural world. Engineering is the application of knowledge in order to design, build and maintain a product or a process that solves a problem and fulfills a need (i.e. a technology).