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Dimas [21]
3 years ago
14

You spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.111 liters per second and

the diameter of the nozzle you hold is 5.79 mm. At what speed does the water exit the nozzle
Physics
2 answers:
wariber [46]3 years ago
8 0

Given Information:  

diameter of the nozzle = d = 5.79 mm = 0.00579 m

flow rate =  0.111 liters/sec

Required Information:  

Velocity = v = ?

Answer:

Velocity = 4.21 m/s

Explanation:

As we know flow rate is given by

Flow rate = Velocity*Area of nozzle

Where

Area of nozzle = πr²

where

r = d/2

r = 0.00579/2

r = 0.002895 m

Area of nozzle = πr²

Area of nozzle = π(0.002895)²

Area of nozzle = 2.6329x10⁻⁵ m²

Velocity = Flow rate/area of nozzle

Divide the litters/s by 1000 to convert into m³/s

0.111/1000 = 1.11x10⁻⁴ m³/s

Velocity = 1.11x10⁻⁴/2.6329x10⁻⁵

Velocity = 4.21 m/s

Therefore, the water exit the nozzle at a speed of 4.21 m/s

Marrrta [24]3 years ago
7 0

Answer:

29.5 m/s

Explanation:

Volumetric flowrate = (average velocity of flow) × (cross sectional area)

Volumetric flowrate = 0.111 liters/s = 0.000111 m³/s

Cross sectional Area of flow = πr²

Diameter = 0.00579 m,

Radius, r = d/2 = 0.002895 m

A = π(0.002895)² = 0.0000037629 m²

Velocity of flow = (volumetric flow rate)/(cross sectional Area of flow)

v = 0.000111/0.0000037629

v = 29.5 m/s

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Explanation:

Given:

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Now, electric field due to charges is given as:

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E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

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<u>Force by each of the charges at the remaining corners:</u>

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E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

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