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Varvara68 [4.7K]
3 years ago
9

Water at 288.7 k flows through a 15.24-cm. diameter horizontal pipe into a 190-l drum. determine the minimum time it will take t

o fill the drum if the flow remains laminar.
Physics
1 answer:
nikdorinn [45]3 years ago
7 0

The flow will be laminar if Reynold's numberN_{R} is less than 2000.

Use the Reynold's formula and rearrange to calculate velocity of water in the pipe.

N_{R} = \frac{v D}{\nu}

Where, v is velocity of the fluid, D is the diameter of pipe, and \nu is the kinematic viscosity i.e. 1.12 \times 10^{-6} m^2/s for water at 288.7 K from Appendix.

So, velocity is:

v = \frac{N_{R} \nu}{D}

The flow rate <em>Q</em>:

Q = vA=v\pi D^2/4=\frac{\frac{N_R \nu}{D} \pi D^2}{4} =[tex] t = \frac{V}{\frac{N_R \nu \pi D}{4}} = \frac{4V}{N_R \nu \pi D} =\frac{4 \times 190 L\frac{10^{-3} m^3}{L}}{2000 \times 1.12\times 10^{-6} m^2/s \pi 15.24 cm\frac{1 m}{100 cm}}  = 2226.3 s \frac{1 min}{60 s}= 37.1 min [/tex]

Where A is the area of cross section of pipe.

The time taken to fill is:

t = Q/V

Where V is the capacity of the tank.


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A hydrometer is made of a tube of diameter 2.3cm.The mass of the tube and it's content is 80g. If it floats in a liquid density
iris [78.8K]

Answer:

The depth to which the hydrometer sinks is approximately 24.07 cm

Explanation:

The given parameters are;

The diameter of the hydrometer tube, d = 2.3 cm

The mass of the content of the tube, m = 80 g

The density of the liquid in which the tube floats, ρ = 800 kg/m³

By Archimedes' principle, the up thrust (buoyancy) force acting on the hydrometer = The weight of the displaced liquid

When the hydrometer floats, the up-thrust is equal to the weight of the hydrometer which by Archimedes' principle, is equal to the weight of the volume of the liquid displaced by the hydrometer

Therefore;

The weight of the liquid displaced = The weight of the hydrometer, W = m·g

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

∴ W = 80 g × g

The volume of the liquid that has a mass of 80 g (0.08 kg), V = m/ρ

V = 0.08 kg/(800 kg/m³) = 0.0001 m³ = 0.0001 m³ × 1 × 10⁶ cm³/m³ = 100 cm³

The volume of the liquid displaced = 100 cm³ = The volume of the hydrometer submerged, V_h

V_h = A × h

Where;

A = The cross-sectional area of the tube = π·d²/4

h = The depth to which the hydrometer sinks

h = V_h/A

∴ h = 100 cm³/( π × 2.3²/4 cm²) ≈ 24.07 cm

The depth to which the tube sinks, h ≈ 24.07 cm.

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What are electric motors?
Lapatulllka [165]
It's an electric motor that converts electric energy into Mechanical energy
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A runner went from 6 m/s and two seconds what was his acceleration
Radda [10]

Answer:

is it 3?

Explanation:

Im taking a guess and just dividing 6 and 2

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Velocity can best be described by which of the following statements?
pogonyaev

Answer: Velocity can best be described as, the speed in a given direction.

Explanation: To find the answer, we need to know more about the Velocity of a body.

<h3>What is Velocity of a body?</h3>
  • Velocity is the rate of change of displacement.
  • It's a vector quantity and is measured in m/s.
  • It can be positive, negative or zero.
  • A body is said to be in uniform motion, then its velocity remains constant.
  • Change in velocity can be a change in speed.
  • The magnitude of velocity is less than or equal to speed.

Thus, we can conclude that, the option C is best describing velocity.

Learn more about velocity here:

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8 0
2 years ago
Read 2 more answers
011 10.0 points
Ulleksa [173]

Answer:

2.47 m

Explanation:

Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.

The horizontal velocity of the ball is constant:

v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s

and the time taken to cover the horizontal distance d is

t=\frac{d}{v_x}=\frac{52}{20.3}=2.56 s

So this is the time the ball takes to reach the horizontal position of the crossbar.

The vertical position of the ball at time t is given by

y=u_y t - \frac{1}{2}gt^2

where

u_y = v sin \theta =(25)(sin 35.9^{\circ})=14.7 m/s is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:

y=(14.7)(2.56) - \frac{1}{2}(9.8)(2.56)^2=5.52 m

The height of the crossbar is h = 3.05 m, so the ball passes

h' = 5.52- 3.05 = 2.47 m

above the crossbar.

8 0
3 years ago
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