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TEA [102]
3 years ago
13

A motion sensor acts as a transmitter when it sends out pulses and as a receiver when it listens for echoes. The delay ∆t betwee

n sending the pulse and receiving the echo is equal to 8.9 ms. The air temperature is equal to 20º C. Part A What is the distance between the motion sensor and the object?
Physics
1 answer:
Oksi-84 [34.3K]3 years ago
4 0

Answer:

1.526 m

Explanation:

The motion sensor sends sound waves in air which travels at a speed of 343 m per second at 20 degree. If d be the distance of the reflecting object

total distance travelled = 2d

Time taken t = 8.9 x 10⁻³ s

velocity v  = 2d / t

343 = 2 d / (8.9 x 10⁻³)

d = 1.526 m

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A permanent magnet is pushed into a wire, left there for a while, and then pulled out. During which time does a current run thou
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Answer:

C. while the magnet is moving

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Electromagnetic induction implies the production of electric current by mere movement of a magnet with respect to a coil or wire.

In the given question, current would be induced in the wire only when the magnet moves. That is either when the magnet is pushed into a wire, or when pulled out. But no current would flow through the wire when the magnet is left there for a while.

The current is induced because of the motion involved. Thus, the appropriate option is C.

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3 years ago
Which of the following types of electromagnetic radiation has the longest wavelength? choose one answer.
Setler79 [48]
The right answer is red light
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Waves that move the particles of the medium parallel to the direction in which the waves are traveling are called a. longitudina
Kamila [148]
Your answer is longitudinal waves. 
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4 years ago
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A portion of the atmosphere that becomes warmer than surrounding air will ____.
earnstyle [38]
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3 0
3 years ago
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
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