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Nesterboy [21]
2 years ago
13

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle conn

ected to a hose can be estimated by the formula where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity. Do you think this equation is valid in any system of units
Physics
1 answer:
suter [353]2 years ago
5 0

This question is incomplete, the complete question is;

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g

where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.

Do you think this equation is valid in any system of units

Answer:

YES, the equation is a general equation that is valid in any system of units

Explanation:

Given the data in the question;

h = (0.04 to 0.09)(D/d)⁴ × \frac{V^{2} }{2g}

so

[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)

[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)

∴ [ L ] = (0.04 to 0.09) [L]

So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.

Therefore, YES, the equation is a general equation that is valid in any system of units

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<h3>What is temperature?</h3>

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5 0
2 years ago
Urgent!
Masja [62]

mass of iron block given as

m_1 = 1.90 kg

density of iron block is

\rho = 7860 kg/m^3

now the volume of the iron piece is given as

V = \frac{m}{\rho}

V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

F_b = \rho_L V g

here we know that

\rho_L = density of liquid = 916 kg/m^3

F_b = 916* 2.42 * 10^{-4} * 9.8

F_b = 2.17 N

Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium

F_s + F_b = mg

F_s + 2.17 = 1.90* 9.8

F_s = 16.45 N

So reading of spring balance will be 16.45 N

Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block

F_n = F_g + F_b

F_n = (1 + 2.50)*9.8 + 2.17

F_n = 34.3 + 2.17 = 36.47 N

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3 0
3 years ago
A suspended object A is attracted to a charged object B, can one conclude that A is charged? Explain
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Explanation:

In my view, when the Object A is attracted to a Charged object B. Object B should be Negatively or Positively charged. So Object B should be the Opposite charged according to the Object B

Example =

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If the Object B is Positively Charged, the Object A should be Negatively Charged

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Answer:

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Explanation:

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To measure the length of her window, she could use an inelastic tape rule or a metre rule. These instruments would eliminate instrumental error.

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