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Semmy [17]
3 years ago
11

What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

hose while carrying a flow of 40.0 L/s? (b) To what maximum height above the nozzle can this water rise? (The actual height will be significantly smaller due to air resistance.)
Physics
1 answer:
Marianna [84]3 years ago
8 0

Answer:

The pressure and maximum height are 1.58\times10^{6}\ N/m^2 and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

We need to calculate the pressure

Using Bernoulli effect

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}

When two point are at same height so ,

P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

Q=A_{1}v_{1}

v_{1}=\dfrac{Q}{A_{1}}

v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}

v_{1}=56.58\ m/s

For output velocity,

v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}

v_{2}=6.28\ m/s

Put the value into the formula

P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)

\Delta P=1.58\times10^{6}\ N/m^2

(b). We need to calculate the maximum height

Using formula of height

\Delta P=\rho g h

Put the value into the formula

1.58\times10^{6}=1000\times9.8\times h

h=\dfrac{1.58\times10^{6}}{1000\times9.8}

h=161.22\ m

Hence, The pressure and maximum height are 1.58\times10^{6}\ N/m^2 and 161.22 m respectively.

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Starting from rest, a 68.0 kg woman jumps down to the floor from a height of 0.790 m, and immediately jumps back up into the air
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Answer:

a) I = 0 N s,  b)  v = -3.935 m / s, c) vf = 3.935 m / s,   d)   y = 0.790 m

Explanation:

a) Let's start by defining the upward direction (+ y) as positive. For this part of the exercise we must use the momentum relationship

          I = ∫F dt

The force of the woman on the floor is given and by action and rection the floor exerts on the woman a force of equal magnitude, but opposite direction

        I = ∫ (9200 t - 11500 t2) dt

        I = 9200 t² / 2 - 11500 t³ / 3

We evaluate between the lower limit t = 0 and upper limit t = 0.800 s

       I = 9200 (0.8² -0) - 11500 (0.8³ -0)

       I = 5888 -5888

       I = 0 N s

Directed from the floor to the woman

b) For this part we use kinematics

      v² = v₀² - 2g y

     v = √ (0 - 2 9.8 (-0.79))

     v = 3.935 m / s

The speed direction is down

c) for this we use the relationship between momentum and the amount of movement

      I = ΔP

      I = m vf - m v₀

     vf = (I + m v₀) / m

This is the impulse of women on the floor    

      vf = ( 0 + 68 (3.935)) / 68

      vf = 3.935 m / s

d) let's use kinematics

      v₂ = v₀² - 2gy

      0 = v₀² - 2gy

      y = v₀² / 2g

      y = 3.935²/2 9.8

      y = 0.790 m

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Which scientist was involved in the War of Currents?<br> Help
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Explanation:

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A circuit contains a single 270-pf capacitor hooked across a battery. it is desired to store four times as much energy in a comb
yan [13]
The energy stored by a system of capacitors is given by
U= \frac{1}{2}C_{eq} V^2
where Ceq is the equivalent capacitance of the system, and V is the voltage applied.

In the formula, we can see there is a direct proportionality between U and C. This means that if we want to increase the energy stored by 4 times, we have to increase C by 4 times, if we keep the same voltage.

Calling C_1 = 270 pF the capacitance of the original capacitor, we can solve the problem by asking that, adding a new capacitor with C_x, the new equivalent capacitance of the system C_{eq} must be equal to 4C_1. If we add the new capacitance X in parallel, the equivalent capacitance of the new system is the sum of the two capacitance
C_{eq} = C_1 + C_x
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A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a
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Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2} (1)

Where:

p_{S,1}, p_{S,2} - Initial and final momemtums of the small object, measured in kilogram-meters per second.

p_{B,1}, p_{B,2} - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that p_{S,1} = 7\,\frac{kg\cdot m}{s}, p_{B,1} = 0\,\frac{kg\cdot m}{s} and p_{S, 2} = 4\,\frac{kg\cdot m}{s}, then the final momentum of the big object is:

7\,\frac{kg\cdot m}{s} + 0\,\frac{kg\cdot m}{s} = 4\,\frac{kg\cdot m}{s}+p_{B,2}

p_{B,2} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is:

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

4 0
2 years ago
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