This problem here is an example of inelastic collision where kinetic energy is not conserved but momentum is. We calculate as follows:
m1v1 + m2v2 = (m1 + m2)v3
v3 = m1v1 + m2v2 / m1 + m2
v3 = (30.2)(1000) + (5000)(0) / (30.2 + 5000)
v3 = 6.00 m/s
Answer:
a) 2.33 m/s
b) 5.21 m/s
c) 882 m³
Explanation:
Using the concept of continuity equation
for flow through pipes
![A_{1}\times V_{1} = A_{2}\times V_{2}](https://tex.z-dn.net/?f=A_%7B1%7D%5Ctimes%20V_%7B1%7D%20%3D%20A_%7B2%7D%5Ctimes%20V_%7B2%7D)
Where,
A = Area of cross-section
V = Velocity of fluid at the particular cross-section
given:
![A_{1} = 0.070 m^{2}](https://tex.z-dn.net/?f=A_%7B1%7D%20%3D%200.070%20m%5E%7B2%7D)
![V_{1} = 3.50 m/s](https://tex.z-dn.net/?f=V_%7B1%7D%20%3D%203.50%20m%2Fs)
a) ![A_{2} = 0.105 m^{2}](https://tex.z-dn.net/?f=A_%7B2%7D%20%3D%200.105%20m%5E%7B2%7D)
substituting the values in the continuity equation, we get
![0.070\times 3.50 = 0.105\times V_{2}](https://tex.z-dn.net/?f=0.070%5Ctimes%203.50%20%3D%200.105%5Ctimes%20V_%7B2%7D)
or
![V_{2} = \frac{0.070\times 3.5}{0.150}m/s](https://tex.z-dn.net/?f=V_%7B2%7D%20%3D%20%5Cfrac%7B0.070%5Ctimes%203.5%7D%7B0.150%7Dm%2Fs)
or
![V_{2} = 2.33m/s](https://tex.z-dn.net/?f=V_%7B2%7D%20%3D%202.33m%2Fs)
b) ![A_{2} = 0.047 m^{2}](https://tex.z-dn.net/?f=A_%7B2%7D%20%3D%200.047%20m%5E%7B2%7D)
substituting the values in the continuity equation, we get
![0.070\times 3.50 = 0.047\times V_{2}](https://tex.z-dn.net/?f=0.070%5Ctimes%203.50%20%3D%200.047%5Ctimes%20V_%7B2%7D)
or
![V_{2} = \frac{0.070\times 3.5}{0.047}m/s](https://tex.z-dn.net/?f=V_%7B2%7D%20%3D%20%5Cfrac%7B0.070%5Ctimes%203.5%7D%7B0.047%7Dm%2Fs)
or
![V_{2} = 5.21m/s](https://tex.z-dn.net/?f=V_%7B2%7D%20%3D%205.21m%2Fs)
c) we have,
Discharge![Q = Area (A)\times Velocity(V)](https://tex.z-dn.net/?f=Q%20%3D%20Area%20%28A%29%5Ctimes%20Velocity%28V%29)
thus from the given value, we get
![Q = 0.070m^{2}\times 3.5m/s\](https://tex.z-dn.net/?f=Q%20%3D%200.070m%5E%7B2%7D%5Ctimes%203.5m%2Fs%5C)
![Q = 0.245 m^{3}/s](https://tex.z-dn.net/?f=Q%20%3D%200.245%20m%5E%7B3%7D%2Fs)
Also,
Discharge![Q = \frac{volume}{time}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7Bvolume%7D%7Btime%7D)
given time = 1 hour = 1 ×3600 seconds
substituting the value of discharge and time in the above equation, we get
![0.245m^{3}/s = \frac{volume}{3600s}](https://tex.z-dn.net/?f=0.245m%5E%7B3%7D%2Fs%20%3D%20%5Cfrac%7Bvolume%7D%7B3600s%7D)
or
![0.245m^{3}/s\times 3600 = Volume](https://tex.z-dn.net/?f=0.245m%5E%7B3%7D%2Fs%5Ctimes%203600%20%3D%20Volume)
volume of flow = ![882 m^{3}](https://tex.z-dn.net/?f=882%20m%5E%7B3%7D)
Answer:
<em>Digital clock is the most precised one</em>
<em></em>
Explanation:
Answer:α = 1.00 rad/s², τ = 90.0 N•m, KEr = 1.12 kJ
Explanation:
m = 795/9.81 = 81.0 kg
ω₁ = 47.79 rev/min(2π rad/rev) / (60 s/min) = 5.00 rad/s
α = (ω₁ - ω₀)/τ = (5.00 - 0.00)/5.00 = 1.00 rad/s²
I = ½mR² = ½(81.0)(1.49²) = 90.0 kg•m²
τ = Iα = 90.0(1.00) = 90.0 N•m
KEr = ½Iω² = ½(90.0)5.00² = 1,124.477 ≈ 1.12 kJ
No, it simply consists of elements.