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weqwewe [10]
3 years ago
7

15. You are watching a baseball game on television that 15

Physics
1 answer:
Gre4nikov [31]3 years ago
5 0

Answer:

t_total = 6.99 s

Explanation:

It asks us how long it takes to hear the sound, for this we must look for the time (t₁) it takes for the sound to reach the microphone, the time it takes for the video signal (t₂) to reach the television and the time (₃) it takes for the TV sound to reach us, so the total delay time is

         t_total = t₁ + t₂ + t₂

we look for t1, it indicates that the distance x = 22m

          v = x / t

           t = x / v

           t₁ = 22/343

           t₁ = 6.41 10-2 s

time t₂

          t₂ = 4500 103/3 108

          t₂ = 1.5 10-5 s

time t₃

          t₃ = 2/343

          t₃ = 5.83 10⁻³

Total time is

         t_total = t₁ + t₂ + t₃

         t_total = 6.41 10⁻² + ​​1.5 10⁻⁵ + 0.583 10⁻²

         t_total = 6.99 s

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Information that is given:
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v0 = 25 m/s
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The velocity at the end of the motion equals zero (0) because the car stops, so v=0.
0 = v0^2 - 2aS
v0^2 = 2aS
S = v0^2/2a
S = (25 m/s)^2/(2×5.4 m/s^2)
S = (25 m/s)^2/(10.8 m/s^2)
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True, the path of the ball, as observed from the train window, will be a horizontal straight line.

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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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