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3 years ago

Periodic motion can BEST be defined as motion

Physics

1 answer:

koban [17]3 years ago

8 0

Answer:

Explanation:

repeated in intervals of time

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Which of the following wavelengths will produce standing waves on a string that is 3.5 m long?

denpristay [2]

In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

$\lambda=\frac{2}{n} L$

The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:

$\lambda=\frac{2}{1} \cdot 3.5 m=7.0 m$

The wavelength of the 2nd harmonic is:

$\lambda=\frac{2}{2} \cdot 3.5 m=3.5 m$

The wavelength of the 4th harmonic is:

$\lambda=\frac{2}{4} \cdot 3.5 m=1.75 m$

It is not possible to find any integer n such that $\lambda=5 m$ , therefore the correct options are A, B and D.

3 0

3 years ago

Read 2 more answers

Four importance of soil water

Andru [333]

Answer:

nitrogen, phosphorus, potassium, and calcium

Explanation:

Just put it.

4 0

2 years ago

True of false efficiency compared the output work to the output force

Lerok [7]

The statement about "<span>efficiency compared the output work to the output force" is false. Efficiency can be compared from the input work to the output work.</span>

8 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

While goofing off at the ice skating rink, a student takes off her shoes and places each of them on the ice. Her friend, a hocke

3241004551 [841]

Answer:

The right shoe

Explanation:

Both shoes have the same speed.

5 0

3 years ago