Answer:
Option B

Explanation:








Centripetal acceleration 
Tangential component=dr=2*1.75=3.5

 
        
             
        
        
        
Why not simply include each step's setup instructions within it? is it due to our desire to maintain DRY (Don't Repeat Yourself) code?
<h3>What is the purpose of unit testing?</h3>
Program testing is known as "unit testing" involves testing individual software components. When developing an application, unit testing is done on the software product. An individual component could be a technique or a specific function.
Unit testing's primary goal is to separate written code for testing to see if it functions as intended. Unit testing is a crucial stage in the development process because, when done properly, it can aid in finding early code issues that could be more challenging to identify in subsequent testing phases.
The core of the testing process consists of unit testing and functional testing. The primary distinction between the two is that during the development cycle, the developer conducts unit testing. The tester does functional testing at the system testing level.
To learn more about unit testing refers to:
brainly.com/question/22900395
#SPJ4
 
        
             
        
        
        
Answer:
the maximum thermal efficiency is 29%
Explanation:
the maximum efficiency for a thermal engine that works between a cold source and a hot source is the one of a Carnot engine. Its efficiency is given by
Maximum efficiency= 1 - T2/T1
where 
T2= absolute temperature of the cold sink (environment)= 20°C + 273 = 293
T2= absolute temperature of the hot source (hot water supply) = 140°C + 273 = 413
therefore
Maximum efficiency= 1 - T2/T1 = 1 - 293/413 = 0,29 =29%
 
        
             
        
        
        
Answer:
The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C
Explanation:
The properties of water at 100°C and 1 atm are:
pL = 957.9 kg/m³
pV = 0.596 kg/m³
ΔHL = 2257 kJ/kg
CpL = 4.217 kJ/kg K
uL = 279x10⁻⁶Ns/m²
KL = 0.68 W/m K
σ = 58.9x10³N/m
When the water boils on the surface its heat flux is:

For copper-water, the properties are:
Cfg = 0.0128
The heat flux is:
qn = 0.9 * 18703.42 = 16833.078 W/m²

The tube surface temperature immediately after installation is:
Tinst = 100 + 20.4 = 120.4°C
For rough surfaces, Cfg = 0.0068. Using the same equation:
ΔT = 10.8°C
The tube surface temperature after prolonged service is:
Tprolo = 100 + 10.8 = 110.8°C
 
        
             
        
        
        
Answer:
D=41.48 ft

Explanation:
Given that
y=0.5 x²                      
Vx= 2 t
We know that

At t= 0 ,x=0   

At t= 3 s

![x=[t^2\left\right ]_0^3](https://tex.z-dn.net/?f=x%3D%5Bt%5E2%5Cleft%5Cright%20%5D_0%5E3)
x= 9 ft
When x= 9 ft then 
y= 0.5 x 9²  ft
y= 40.5 ft
So distance from origin is
x= 9 ft ,y= 40.5 ft

D=41.48 ft

Vx= 2 t

At t= 3 s , x= 9 ft 
y=0.5 x²    

y=0.5 x²    


Given that







