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max2010maxim [7]
3 years ago
8

A thin plate 6 ft long and 3 ft wide is submerged and held stationary in a stream of water (T = 60°F) that has a velocity of 17

ft/s. What is the thickness of the boundary layer on the plate for Rex = 500,000 (assume that the boundary layer is still laminar), and at what distance downstream of the leading edge does this Reynolds number occur? What is the shear stress on the plate at this point?
Physics
1 answer:
VashaNatasha [74]3 years ago
4 0

A) In the case of the Boundary Thickness Layer we use the given formula,

\delta = \frac{4.91x}{\sqrt{Re}}

We know as well that,

Re = Número de Reynolds = \frac{U*x}{\upsilon}

Where,

U = velocity

\upsilon = kinematic viscosity

For water, kinematic viscosity, \upsilon = 1.21*10^{-5} ft^2 /s

So, 500,000 = \frac{ 17x}{(1.21*10^{-5})}

x = 0.355 ft

d = \frac{4.91*0.355}{\sqrt {500000}}

d = 0.002465 ft = 0.029in

B) For flat plate boundary layer. Given the Critical Reynolds Number.= 5*10^5 we know that is equal to Re above.

Thus, x = 0.355 ft

C. Wall shear stress,

\tau = \mu*\sqrt{ U^3 / (2*\nu*x) }

For water, dynamic viscosity, \nu = 2.344*10^-5 lbf-s/ft^2

\tau = 2.344*10^-5 \sqrt {17^3 / (2*1.21*10^{-5}*0.355)}

\tau = 0.5605 lbf/ft^2

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rusak2 [61]

Answer:

Spring constant in N / m = 6,000

Explanation:

Given:

Length of spring stretches = 5 cm = 0.05 m

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Find:

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Spring constant in N / m = Force/Distance

Spring constant in N / m = 300 / 0.05

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3 years ago
A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most
Oxana [17]

Answer:

A)  I_{total} = 1.44 kg m², B) moment of inertia must increase

Explanation:

The moment of inertia is defined by

     I = ∫ r² dm

For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is

      I = ½ m R²

A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

    I = I_{cm} + m D²

Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

      I_{total}=I_{body} + 2 I_{arm}

       I_{body} = ½ M R²

The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

       M = 7/8 m total

       M = 7/8 64

       M = 56 kg

The mass of the arms is

      m’= 1/8 m total

      m’= 1/8 64

      m’= 8 kg

As it has two arms the mass of each arm is half

     m = ½ m ’

     m = 4 kg

The arms are very thin, we will approximate them as a particle

    I_{arm} = M D²

Let's write the equation

     I_{total} = ½ M R² + 2 (m D²)

Let's calculate

    I_{total} = ½ 56 0.20² + 2 4 0.20²

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b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

6 0
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What do astronomers use in addition to parallax to find the actual distance of stars that are close to Earth?
IgorLugansk [536]

Answer:

trigonometry (guessing)

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ellipse: is the shape of an orbit : looks like an oval

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https://science.howstuffworks.com/question224.htm

By comparing the intrinsic brightness to the star's apparent brightness we can calculate the distance of stars

1/r^2 rule states that the apparent brightness of a light source is proportional to the square of its distance.Jan 11, 2022

https://www.space.com/30417-parallax.html

alternative distance measurement for stars used by most astronomers is the parsec. A star with a parallax angle of 1 arcsecond has a distance of 1 parsec, or 1 parsec per arcsecond of parallax, which is about 3.26 light years

blossoms.mit.edu

.

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