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cluponka [151]
3 years ago
9

Water flows with constant speed through a garden hose that goes up to 27.5 cm high. if the water pressure is 132kpa at the botto

m of the step, what is its pressure at the top of the step?
Physics
1 answer:
sergejj [24]3 years ago
6 0

Answer:

The pressure at the top of the step is 129.303 kilopascals.

Explanation:

From Hydrostatics we find that the pressure difference between extremes of the water column is defined by the following formula, which is a particular case of the Bernoulli's Principle (v_{bottom}\approx v_{top}):

p_{bottom}-p_{top} = \rho\cdot g\cdot \Delta h (1)

p_{bottom}, p_{top} - Total pressures at the bottom and at the top, measured in pascals.

\rho - Density of the water, measured in kilograms per cubic meter.

\Delta h - Height difference of the step, measured in meters.

If we know that p_{bottom} = 132000\,Pa, \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}} and \Delta h = 0.275\,m, then the pressure at the top of the step is:

p_{top} = p_{bottom}-\rho\cdot g\cdot \Delta h

p_{top} = 132000\,Pa-\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.275\,m)

p_{top} = 129303.075\,Pa

p_{top} = 129.303\,kPa

The pressure at the top of the step is 129.303 kilopascals.

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One of your classmates, in a fit of unrestrained ego, jumps onto a lab table:
Anettt [7]

The equilibrium condition allows finding the results for the forces of the system are

      a) The free body diagram is in the attachment

      b) The normal force is N = 737 N

      c) The mass of the table is  10.2 kg

Newton's second law indicates that the net force is proportional to the product of the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition

          ∑ F = 0

Where the bold letters indicate vectors, F is the external forces

a) A free body diagram is a scheme of the forces without the details of the bodies, in the attachmentt we see a free body diagram of the system.

b) The reaction force of the ground is applied in each of the legs of the table, in general this force has the same magnitude in each leg, therefore in Newton's second law we can place it as a single force

             N = N₁ + N₂ + N₃ + N₄₄

Let's apply the equilibrium condition

                N -  W_m -w_{table} = 0

                N =  W_m +w_{table}

                N = M_m g + w_{table}  

They indicate the pose of the boy is 65 kg, for the weight of the table of a laboratory table is approximately 100 N

                N = 65 9.8 + 100

                N = 737 N

c) To calculate the mass of the table we use the relation

                W = m_{table} g

                m_{table} = \frac{w_{table}}{g}

                m_{tabble}= \frac{100}{9.8}  

               m_{table}e = 10.2 kg

In conclusion using the equilibrium condition we can find the results for the forces are

      a) The free body diagram is in the attachment

      b) The normal force is N = 737 N

      c) The mass of the table is  10.2 kg

Learn more here:  brainly.com/question/19860811

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Answer:

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Explanation:

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