Answer:
Explanation:
Given
Volume of bucket ![V=10\ gallon](https://tex.z-dn.net/?f=V%3D10%5C%20gallon)
Time taken to fill the bucket ![t=50\ s](https://tex.z-dn.net/?f=t%3D50%5C%20s)
so volume flow rate is ![\dot{V}=\frac{10}{50}=0.2\ gal/s](https://tex.z-dn.net/?f=%5Cdot%7BV%7D%3D%5Cfrac%7B10%7D%7B50%7D%3D0.2%5C%20gal%2Fs)
1 gal is equivalent to ![0.133\ ft^3](https://tex.z-dn.net/?f=0.133%5C%20ft%5E3)
![\dot{V}=0.0267\ ft^3/s](https://tex.z-dn.net/?f=%5Cdot%7BV%7D%3D0.0267%5C%20ft%5E3%2Fs)
mass flow rate ![\dot{m}=\rho \times \dot{V}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%3D%5Crho%20%5Ctimes%20%5Cdot%7BV%7D)
![\dot{m}=62.4\times 0.0267](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%3D62.4%5Ctimes%200.0267)
![\dot{m}=1.668\ lbs](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%3D1.668%5C%20lbs)
(b)Average velocity through nozzle exit
![\dot{V}=Av_{avg}](https://tex.z-dn.net/?f=%5Cdot%7BV%7D%3DAv_%7Bavg%7D)
![v_{avg}=\dfrac{0.0267}{\frac{\pi}{4}\times (0.0262)^2}](https://tex.z-dn.net/?f=v_%7Bavg%7D%3D%5Cdfrac%7B0.0267%7D%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%5Ctimes%20%280.0262%29%5E2%7D)
![v_{avg}=49.51\ ft/s](https://tex.z-dn.net/?f=v_%7Bavg%7D%3D49.51%5C%20ft%2Fs)
Answer:571.09 kJ
Explanation:
Given
Temperature of cooling water from engine exit![=240^{\circ} F\approx 115.55^{\circ}C](https://tex.z-dn.net/?f=%3D240%5E%7B%5Ccirc%7D%20F%5Capprox%20115.55%5E%7B%5Ccirc%7DC)
After Passing through the radiator its temperature decreases to ![175^{\circ}F\approx 79.44^{\circ}F](https://tex.z-dn.net/?f=175%5E%7B%5Ccirc%7DF%5Capprox%2079.44%5E%7B%5Ccirc%7DF)
specific heat of water![=4.184 J/g^{\circ}C](https://tex.z-dn.net/?f=%3D4.184%20J%2Fg%5E%7B%5Ccirc%7DC)
Volume of water ![= 1 gallon\approx 3.78 L](https://tex.z-dn.net/?f=%3D%201%20gallon%5Capprox%203.78%20L)
density of water ![\rho =1 gm/mL](https://tex.z-dn.net/?f=%5Crho%20%3D1%20gm%2FmL)
Thus mass of water![=\rho \times V=3.78\times 1=3.78 kg](https://tex.z-dn.net/?f=%3D%5Crho%20%5Ctimes%20V%3D3.78%5Ctimes%201%3D3.78%20kg%20)
Heat transferred to the surrounding is equal to heat absorbed by cooling water
![Q=m\cdot c\cdot \Delta T](https://tex.z-dn.net/?f=Q%3Dm%5Ccdot%20c%5Ccdot%20%5CDelta%20T)
![Q=3.78\times 4.184\times 1000\times (115.55-79.44)](https://tex.z-dn.net/?f=Q%3D3.78%5Ctimes%204.184%5Ctimes%201000%5Ctimes%20%28115.55-79.44%29)
![Q=3.78\times 4.184\times 1000\times (36.11)](https://tex.z-dn.net/?f=Q%3D3.78%5Ctimes%204.184%5Ctimes%201000%5Ctimes%20%2836.11%29)
![Q=571.09 kJ](https://tex.z-dn.net/?f=Q%3D571.09%20kJ)
Answer:
The answer is below
Explanation:
a) The change in energy is the difference between the final energy and the initial energy.
ΔE (energy change) = Ef (final energy) - Ei (initial energy)
![\Delta E=\frac{1}{2}kA_f^2 -\frac{1}{2}kA_i^2\\\\k=force\ constant=2.7\ N/cm=270\ N/m, A_f=final\ dispalacment= 3.7\ cm=0.037\ m,\\ A_i=initial \ displacement = 6.2\ cm=0.062\ m\\\\Hence:\\\\\Delta E=\frac{1}{2}(270)(0.037)^2 -\frac{1}{2}(270)(0.062)^2\\\\\Delta E=-0.334 \ J](https://tex.z-dn.net/?f=%5CDelta%20E%3D%5Cfrac%7B1%7D%7B2%7DkA_f%5E2%20-%5Cfrac%7B1%7D%7B2%7DkA_i%5E2%5C%5C%5C%5Ck%3Dforce%5C%20constant%3D2.7%5C%20N%2Fcm%3D270%5C%20N%2Fm%2C%20A_f%3Dfinal%5C%20dispalacment%3D%203.7%5C%20cm%3D0.037%5C%20m%2C%5C%5C%20A_i%3Dinitial%20%5C%20displacement%20%3D%206.2%5C%20cm%3D0.062%5C%20m%5C%5C%5C%5CHence%3A%5C%5C%5C%5C%5CDelta%20E%3D%5Cfrac%7B1%7D%7B2%7D%28270%29%280.037%29%5E2%20-%5Cfrac%7B1%7D%7B2%7D%28270%29%280.062%29%5E2%5C%5C%5C%5C%5CDelta%20E%3D-0.334%20%5C%20J)
The negative sign shows that energy is lost to the environment. Hence 0.334 J is lost to the environment.
b) According to the law of conservation of energy, energy cannot be created or destroyed but transformed from one form to another.
The oscillating object loses energy due to wind resistance, friction between the spring and the object. Given that the air is frictionless, hence the energy loss is due to friction which is converted to heat.
Answer:
A , power
Explanation:
Hope this is useful. Have a lovely rest of your day! God bless you.