Information gathered from observing a plant grow 3 cm over a two week period is<br> called:

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

3 years ago

A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA. What amount of vars must be added to bring the pf to 0.85

kvasek [131]

Answer:

$\mathtt{Q_{sh} = 600.75 \ vars}$

Explanation:

Given that:

A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA

Here:

the initial power factor i.e cos θ₁ = 0.7 lag

θ₁ = cos⁻¹ (0.7)

θ₁ = 45.573°

Active power P = 1500 watts

Apparent power S = 2100 VA

What amount of vars must be added to bring the pf to 0.85

i.e the required power factor here is cos θ₂ = 0.85 lag

θ₂ = cos⁻¹ (0.85)

θ₂ = 31.788°

However; the initial reactive power $Q_1$ = P×tanθ₁

the initial reactive power $Q_1$ = 1500 × tan(45.573)

the initial reactive power $Q_1$ = 1500 × 1.0202

the initial reactive power $Q_1$ = 1530.3 vars

The amount of vars that must therefore be added to bring the pf to 0.85

can be calculated as:

$Q_{sh} = P( tan \theta_1 - tan \theta_2)$

$Q_{sh} = 1500( tan \ 45.573 - tan \ 31.788)$

$Q_{sh} = 1500( 1.0202 - 0.6197)$

$Q_{sh} = 1500( 0.4005)$

$\mathtt{Q_{sh} = 600.75 \ vars}$

3 0

3 years ago

A force that tries to slow things down when two things are rubbed together

irakobra [83]

Answer:

Friction

Explanation:

friction is the force that tries to slow things down when two things are rubbed together.

5 0

3 years ago

Read 2 more answers

An object is swung in a horizontal circle on a length of string that is 0.93 m long. Its acceleration is 26.36 m/s2. What is the

Igoryamba

Answer:

The object takes approximately 1.180 seconds to complete one horizontal circle.

Explanation:

From statement we know that the object is experimenting an Uniform Circular Motion, in which acceleration ( $a$ ), measured in meters per square second, is entirely centripetal and is expressed as:

$a = \frac{4\pi^{2}\cdot R}{T^{2}}$ (1)

Where:

$T$ - Period of rotation, measured in seconds.

$R$ - Radius of rotation, measured in meters.

If we know that $a = 26.36\,\frac{m}{s^{2}}$ and $R = 0.93\,m$ , then the time taken by the object to complete one revolution is:

$T^{2} = \frac{4\pi^{2}\cdot R}{a}$

$T = 2\pi\cdot \sqrt{\frac{R}{a} }$

$T = 2\pi\cdot \sqrt{\frac{0.93\,m}{26.36\,\frac{m}{s^{2}} } }$

$T \approx 1.180\,s$

The object takes approximately 1.180 seconds to complete one horizontal circle.

7 0

3 years ago

Which statements describe magnetic poles? Check all that apply.

Alex_Xolod [135]

Explanation:

Magnet: It has two poles: South pole and North pole.

Magnetic field lines are stronger near the poles of the magnet.

Same poles repel each other. There is a magnetic force of repulsion between the same poles. North- North poles repel each other.

Unlike poles attract each other. There is magnetic force of attraction between the opposite poles. South- North poles attract each other.

Mono poles cannot exist.

From the given statements, the magnetic poles are described by:

A north pole must exist with a south pole.

Two south poles placed near each other will repel each other.

A north pole and a south pole placed near each other will attract each other.

5 0

3 years ago

Read 2 more answers

Information gathered from observing a plant grow 3 cm over a two week period is called: