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Sonbull [250]
2 years ago
12

Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous

value. If the speed of the water in the larger section of the pipe was what is its speed in this smaller section if the water behaves like an ideal incompressible fluid
Physics
1 answer:
Orlov [11]2 years ago
3 0

Answer:

Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

Continuity equation is applicable for ideal, incompressible liquids

Q the flux of water that is  Av with A the cross section area and v the velocity,

so,

A_1V_1=A_2V_2

A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}

the diameter decreases 86% so

d_2 = 0.86d_1

v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}

Thus, speed in smaller section is 48.6 m/s

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A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or
Svetlanka [38]

<u>Answer:</u>

 Mass of base ball  m_b=3.992*10^{14}kg  

<u>Explanation:</u>

  Circumference of baseball = 2πr = 23 cm

  So radius of baseball = 3.66 cm = 3.66*10^{-2} m

   Mass per unit volume of baseball = Mass per unit volume of neutron or proton.

   Mass of proton = 10^{-27} kg  

   Diameter of proton = 10^{-15} m

   Radius of proton =  5*10^{-16} m

   Volume of ball = \frac{4}{3} \pi r^3

   Now substituting all values in Mass per unit volume of baseball = Mass per unit volume of neutron or proton.    

         \frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}

         \frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}  

         m_b=3.992*10^{14}kg

       So mass of base ball  m_b=3.992*10^{14}kg              

5 0
2 years ago
Read 2 more answers
A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float
storchak [24]

Answer:

 t ’= \frac{1450}{0.6499 + 2 v_r},  v_r = 1 m/s       t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

        v_b - v_r = d / t

By the time the boat goes down the river

        v_b + v_r = d '/ t'

let's subtract the equations

       2 v_r = d ’/ t’ - d / t

       d ’/ t’ = 2v_r + d / t

       t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }

In the exercise they tell us

         d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

         d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

      t ’= \frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \  + 2 \ v_r}

      t ’= \frac{1450}{0.6499 + 2 v_r}

In order to complete the calculation, we must assume a river speed

          v_r = 1 m / s

       

let's calculate

      t ’= \frac{ 1450}{ 0.6499 + 2 \ 1}

      t ’= 547.19 s

8 0
2 years ago
A frictionless pendulum clock on the surface of the earth has a period of 1.00 s. On a distant planet, the length of the pendulu
Sergio [31]

Answer:

The gravitational acceleration on the planet is slightly less than g.

Explanation:

The period of a pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity

The formula can also be rewritten as

g=(\frac{2\pi}{T})^2 L (1)

In this problem, we have a pendulum which has a period of T=1.00 s on Earth. The length of the same pendulum must be shortened on the distant planet to have the same period of T'=1.00 s: this means that the length of the pendulum on the distant planet, L', is shorter than the length of the pendulum on Earth, L

L'

By looking at formula (1), we see that g (the gravitational acceleration) is directly proportional to L. therefore, if L is shortened on the distant planet, it means that also the value of g is lower than on Earth:

so, the correct answer is

The gravitational acceleration on the planet is slightly less than g.

7 0
2 years ago
A refrigerator with a weight of 1,127 newtons needs to be moved into a house using a ramp. The length of the ramp is 2.1 meters,
lina2011 [118]
496/1127 = 0.44 = 44% 

<span>sin A = 0.85/2.1. </span>
<span>A = 23.9o. </span>

<span>Fp = 1127 sin23.9 = 457 N. = Force parallel to the ramp. </span>

<span>Fn = 1127 Cos23.9 = 1,030 N. = Force </span>
<span>perpendicular to the ramp = Normal force. </span>

<span>Eff. = Fp/Fap = 457/496 = 0.92 = 92%
Correct answer: 92%</span>
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liberstina [14]
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5 0
2 years ago
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