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Andrej [43]
3 years ago
13

Absolute pressure in tank is P1 = 260 kPa and local ambient absolute pressure is P2 =100 kPa. If liquid density in pipe is 13600

kg/m3 , compute liquid height, h=..?.. m ? Use g =10 m/s2
Physics
1 answer:
Pavlova-9 [17]3 years ago
5 0

Answer:

1.176m

Explanation:

Local ambient pressure(P1) = 100 kPa

Absolute pressure(P2)=260kPa

Net pressure=absolute pressure-local ambient absolute pressure

Net pressure=P1(absolute pressure)-P2(local ambient absolute pressure)

Net pressure=260-100=160kPa

Pressure= ρgh

160kPa=13600*10*h

h=\frac{160000}{136000}

h=1.176m

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Answer:

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x = v0*t + (a*t²/2)

x = 21.5t + (6*t²/2)

x = 21.5t + 3t²   <em>(I)</em>

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x = 52 + 21.5t  <em>(II)</em>

<em />

then we can apply <em>I = II</em>

21.5t + 3t² = 52 + 21.5t

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At the intersection of Texas Avenue and University Drive,
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Answer:

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We can use conservation of momentum to find the initial velocities.

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\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )

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\vec{P}_f = (2.850 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

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As there are not external forces, the total linear momentum must be constant.

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( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )  

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\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}  } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.

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m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

v_{truck} = 21.93 \frac{m}{s}

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v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}

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