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2 years ago

State and briefly describe the applications of expansion

1 answer:

6 0

When solid material expands in response to an increase in temperature (thermal expansion), it can increase in length in a process known as linear expansion. for an example application of expansion and contraction.

examples =
(1) Changing of shape and dimensions of objects such as doors.
(2) Wall collapsing due to bulging.
(3) Cracking of glass tumbler due to heating.
(4) Bursting of metal pipes carrying hot water or steam are some of the disadvantages of thermal expansion of matter.

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Two runners are competing in track meet. Samantha runs with a speed of 10 m/s and completes her race in 10 seconds. Jordin takes

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The answer for this question is B.

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3 years ago

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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:

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

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3 years ago

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san4es73 [151]

Answer:

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Explanation:

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3 years ago

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Light in the air is incident at an angle to the surface of (12.0 A) degrees on a piece of glass with an index of refraction of (

Orlov [11]

The question is incomplete. You dis not provide values for A and B. Here is the complete question

Light in the air is incident at an angle to a surface of (12.0 + A) degrees on a piece of glass with an index of refraction of (1.10 + (B/100)). What is the angle between the surface and the light ray once in the glass? Give your answer in degrees and rounded to three significant figures.

A = 12

B = 18

Answer:

18.5⁰

Explanation:

Angle of incidence i = 12.0 + A

A = 12

= 12.0 + 12

= 14

Refractive index u = 1.10 + B/100

= 1.10 + 18/100

= 1.10 + 0.18

= 1.28

We then find the angle of refraction index u

u = sine i / sin r

u = sine24/sinr

1.28 = sine 24 / sine r

1.28Sine r = sin24

1.28 sine r = 0.4067

Sine r = 0.4067/1.28

r = sine^-1(0.317)

r = 18.481

= 18.5⁰

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3 years ago

The platform height for Olympic divers is 10 m. A 60 kg diver steps off the platform to begin his dive.

azamat

Answer:

a) Ep = 5886[J]; b) v = 14[m/s]; c) W = 5886[J]; d) F = 1763.4[N]

Explanation:

The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.

$E_{p} =m*g*h\\where:\\m = mass = 60[kg]\\g = gravity = 9.81[m/s^2]\\h = elevation = 10 [m]\\E_{p}=60*9.81*10\\E_{p}=5886[J]$

Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.

$E_{p} = E_{k} \\5886 =0.5*m*v^{2} \\v = \sqrt{\frac{5886}{0.5*60} }\\v = 14[m/s]$

The work is equal to

W = 5886 [J]

We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.

$v_{f} ^{2}= v_{o} ^{2}-2*a*d\\where:\\d = 2.5[m]\\v_{f}=0\\v_{o} =14[m/s]\\Therefore\\a = \frac{14^{2} }{2*2.5} \\a = 39.2[m/s^2]$

By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.

m*g - F = m*a

F = m*a - m*g

F = (60*39.2) - (60*9.81)

F = 1763.4 [N]

3 0

3 years ago

State and briefly describe the applications of expansion​

State and briefly describe the applications of expansion