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masha68 [24]
3 years ago
13

Two pipes are smoothly connected together without leaks. One has a diameter of 3 cm and the other has a diameter of 5 cm. Water

flows through the pipes. In which pipe is the flow speed the greatest? in which pipe is the volume flow rate the greatest?
Physics
1 answer:
Volgvan3 years ago
6 0

Answer:

  1. The speed will be the greatest in the smaller pipe
  2. The volume flow rate is the same in both pipes

Explanation:

This question can be answered using the equation of continuity. Which is

V = A1v1 = A2v2

where V is the volume flow rate

           A is the cross sectional area of the pipe at that position

           v is the velocity of the flow at that point.

From the equation we can see the volume flow rate does not change even as the area and flow speed change. Which is consistent with the assumptions used in this equation which is that

  • The tube must one entry and one exit.
  • The fluid in question is non-viscous.
  • The fluid is incompressible.
  • The flow is steady.  

If the temperature does not change, meaning there is not expansion/contraction and it is incompressible then the volume flow rate does not change.

And the speed will be higher in the pipe with the smaller diameter because area and speed are inversely related.

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A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent
Julli [10]

Answer:

The kinetic energy of the merry-go-round is \bf{475.47~J}.

Explanation:

Given:

Weight of the merry-go-round, W_{g} = 826~N

Radius of the merry-go-round, r = 1.17~m

the force on the merry-go-round, F = 57.8~N

Acceleration due to gravity, g= 9.8~m.s^{-2}

Time given, t=3.47~s

Mass of the merry-go-round is given by

m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg

Moment of inertial of the merry-go-round is given by

I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}

Torque on the merry-go-round is given by

\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m

The angular acceleration is given by

\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}

The angular velocity is given by

\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}

The kinetic energy of the merry-go-round is given by

E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J

5 0
3 years ago
An object accelerates from rest and travels 53 m west in 5.2 s. Determine the acceleration
zmey [24]

Answer:

20.4m/s²

Explanation:

Given parameters:

Initial velocity  = 0m/s

Distance  = 53m

Time  = 5.2s

Unknown:

Acceleration  = ?

Solution:

This is a linear motion and we use the right motion equation;

        S = ut  + \frac{1}{2}at²

S is the distance

u is the initial velocity

a is the acceleration

t is the time

 Insert the parameters and solve;

       53  = (0x 5.2) +  \frac{1}{2} x a x 5.2

       53  = 2.6a

         a = \frac{53}{2.6}  = 20.4m/s²

6 0
2 years ago
Identify each part of this chemical equation that describes the burning of methane and oxygen. B (blue box): D (number): E (purp
Allushta [10]

Answer:

The correct answer is -

A (the entire green box): Chemical Equation

B (the blue box): Reactants

C (the arrow): Reacts to Form

D (the number): Coefficient

E (the purple box): Products

Explanation:

The chemical reaction of burning methane and oxygen is as follows;

Here, the green part A is the chemical equation that includes various parts that are reactants B, methane, and oxygen, C is an arrow that indicates the formation of products.

2 is here coefficient that indicates the moles of the oxygen which forms carbon dioxide and water in box E is products

5 0
2 years ago
It takes four hydrogen nuclei to create one helium nucleus in the proton–proton chain, which is the main energy source of the Su
Artyom0805 [142]

Answer:

0.1371 * 10 ^ -27 kg

Explanation:

From the question number of hydrogen nuclei used to form 3 Helium nuclei = 12 ( 4 * 3 )

mass of the 12 hydrogen nuclei = 12 *( 1.6726 *10^-27) = 20.0712 * 10 ^-27

mass of single helium = 6.6447 * 10 ^-27 kg

therefore the mass of the 3 helium = 3 *( 6.6447 *10 ^-27) = 19.9341 * 10 ^-27 kg

The mass difference between the hydrogen and the helium used

= (20.0712 * 10^ -27)  - (19.9341 * 10 ^-27) = 0.1371 * 10 ^ -27 kg

therefore 0.1371 * 10^-27 kg  is converted into energy to create the three helium nuclei

4 0
2 years ago
Stars and planets are made from gases in a
Law Incorporation [45]
The birthplace of stars are massive cosmic structures formed from gases, also known as nebulas.
3 0
3 years ago
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