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kobusy [5.1K]
3 years ago
9

I will Brainlist "Burning and Inch". Describe this measurement.

Engineering
2 answers:
MAVERICK [17]3 years ago
8 0

Answer:

what measurement

Explanation:

Paladinen [302]3 years ago
4 0

Answer:

You measured from the 1" mark (“burning” the first inch), then forgot to add an inch when marking.

Explanation:

Another way to say it is "Burn an Inch."

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The fan blades suddenly experience an angular acceleration of 2 rad/s2. If the blades are rotating with an initial angular veloc
madreJ [45]

Answer:

Option B

116 ft/s^{2}

Explanation:

\theta=2 rev=2(2\pi)=4\pi

\alpha \theta=0.5(\omega_f^{2}-\omega_i^{2})

\alpha (4\pi)= 0.5(\omega_f^{2}-\omega_i^{2})

\alpha (8\pi)= (\omega_f^{2}-\omega_i^{2})

(2) (8\pi)= (\omega_f^{2}-\omega_i^{2})

(2) (8\pi)= (\omega_f^{2}-4^{2})

\omega_f=8.14 rads/s

v=r\omega=1.75*8.14=14.245 ft/s

Centripetal acceleration =\omega_f^{2} r=8.14^{2}*1.75=115.95 ft/s^{2}

Tangential component=dr=2*1.75=3.5

Resultant=\sqrt{3.5^{2}+115.95^{2}}\approx 116 ft/s^{2}

5 0
3 years ago
Why not just put all the set up steps within each step? it is because we want to keep our code __ ? (3 letters)
andrezito [222]

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

8 0
1 year ago
A well-established way of power generation involves the utilization of geothermal energy-the energy of hot water that exists nat
jeka94

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%

3 0
3 years ago
Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
masya89 [10]

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:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

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

8 0
3 years ago
The particle travels along the path defined by the parabola y=0.5x2, where x and y are in ft. If the component of velocity along
JulsSmile [24]

Answer:

D=41.48 ft

a=54.43\ ft/s^2

Explanation:

Given that

y=0.5 x²                      

Vx= 2 t

We know that

V_x=\dfrac{dx}{dt}

At t= 0 ,x=0  

x=\int V_x.dt

At t= 3 s

x=\int_{0}^{3} 2t.dt

x=[t^2\left\right ]_0^3

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=\sqrt{9^2+40.5^2} \ ft

D=41.48 ft

a_x=\dfrac{dV_x}{dt}

Vx= 2 t

a_x= 2\ ft/s^2

At t= 3 s , x= 9 ft

y=0.5 x²    

a_y=\dfrac{d^2y}{dt^2}

y=0.5 x²    

\dfrac{dy}{dt}=x\dfrac{dx}{dt}

\dfrac{d^2y}{dt^2}=\left(\dfrac{dx}{dt}\right)^2+x\dfrac{d^2x}{dt^2}

Given that

\dfrac{dx}{dt}=2t

\dfrac{dx}{dt}=2\times 3

\dfrac{dx}{dt}=6\ ft/s

a_y=\dfrac{d^2y}{dt^2}=6^2+9\times 2\ ft/s^2

a_y=54\ ft/s^2

a=\sqrt{a_x^2+a_y^2}\ ft/s^2

a=\sqrt{2^2+54^2}\ ft/s^2

a=54.43\ ft/s^2

7 0
3 years ago
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