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3 years ago

If a wave had a wave speed of 1000 m/s and a frequency of 500 Hz, what is its wavelength?

Physics

1 answer:

Reil [10]3 years ago

7 0

Answer:

<h2>2 meters</h2>

Explanation:

<h2>Wavelength = Speed/Frequency </h2><h2>1000 m/s ÷ 500 hz </h2><h2>2 m</h2><h2>hz = s</h2><h2>Hopes this helps. Mark as brainlest plz!</h2>

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Abcdefghijklmnopqrstuvwxyz

AlladinOne [14]

Answer:

lhgwljvqlivlajvliavpjavphvalhvqlhvqlhv

Explanation:

wlhfsougspuva???

6 0

3 years ago

Read 2 more answers

A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a

Makovka662 [10]

Answer:

0.51 m

Explanation:

Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.

Kinetic energy, KE=½kx²

Where k is spring constant and x is the compression of spring

Potential energy, PE=mgh

Where g is acceleration due to gravity, h is height and m is mass

Equating KE=PE

mgh=½kx²

Making x the subject of formula

$x=\sqrt {\frac {2mgh}{k}}$

Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then

$x=\sqrt \frac {2*1300*9.81*10}{1000000}=0.50503465227646m\\x\approx 0.51 m$

5 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

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

Dmitrij [34]

Answer:

$\lambda_2 = 573.3 nm$

Explanation:

As we know that the position of maximum intensity on the screen is given as

$y = \frac{N\lambda L}{d}$

here we know that

$\lambda$ = wavelength

L = distance of the screen

d = distance between two slits

now we know that the position of 8th maximum intensity is same as that of 9th maximum on the screen

so we have

$\frac{N_1\lambda_1 L}{d} = \frac{N_2 \lambda_2 L}{d}$

so here we have

$8 (645 nm) = 9 \lambda_2$

$\lambda_2 = 573.3 nm$

4 0

4 years ago

1. Where is the water in the pot going?

Citrus2011 [14]

Answer:

Explanation:

3 0

3 years ago