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3 years ago

What is the speed of a wave with a frequency of 100 hz and a wave length of .5 m?

Physics

2 answers:

____ [38]3 years ago

7 0

Answer:

Explanation:

There are 2 ways to help with this. Explain the details, which are fairly simple in this topic, or give the formula. My hope is that an explanation will last longer than memorizing the formula. I give you both.

If a wave has frequency, f, of 3 Hz, its period, T, is

. The wavelength,

, is 5 meters. That means that in the time of one period, the wave travels 5 m.

In general,

distance

time

In applying this general definition of speed

↑

to a wave, we have

speed of the wave

wavelength

period

Note: we generally use v for speed of a wave. Using the variable names, then that last formula is written

Since

, we can also say that

⋅

So, using that last formula

⋅

Note: the unit Hz is equivalent to what it was called 100 years ago,

cycles

second

(

also cps

)

. Cycles is not a true unit, so the Hz contributed only the "per second" to the result

ElenaW [278]3 years ago

4 0

Answer: the speed is =30ms^-1 The speed of a wave is given by "speed" (ms^-1)= "frequency(Hz)" xx "wavelength (m)" The frequency is f=100Hz The wavelength is lambda=0.3m The speed is v=lambdaf=0.3*100=30ms^-1

Explanation:

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2 years ago

A table exerts a 4.0 Newton force on a book which lies at rest on its top. The force exerted by the book on the table is

SOVA2 [1]

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3 years ago

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2 years ago

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A 350-g baseball is shot out of a small cannon with a velocity of 9.0 m/s. The baseball flies horizontally at a constant height

stira [4]

Answer:

$9.5 kg m^2/s$

Explanation:

The angular momentum of an object is given by:

$L=mvr$

where

m is the mass of the object

v is its velocity

r is the distance of the object from axis of rotation

Here we have:

m = 350 g = 0.35 kg is the mass of the ball

v = 9.0 m/s is the velocity

r = 3.0 m is the distance of the object from axis of rotation (if we take the ground as the centre of rotation)

Therefore, the angular momentum is:

$L=(0.35)(9.0)(3.0)=9.5 kg m^2/s$

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3 years ago

Two point charges 3q and −8q (with q > 0) are at x = 0 and x = L, respectively, and free to move. A third charge is placed so

riadik2000 [5.3K]

Answer:

Explanation:

The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.

So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.

force on it due to rest of the charges will be equal and opposite so

k3q Q / x² =k 8q Q / (L+x)²

8x² = 3 (L+x)²

2√2 x = √3 (L+x)

2√2 x - √3 x = √3 L

x(2√2 - √3 ) = √3 L

x = √3 L / (2√2 - √3 )

Let us consider the balancing force on 3q

force on it due to -Q and -8q will be equal

kQ . 3q / x² = k3q 8q / L²

Q = 8q (x² / L²)

so charge required = - 8q (x² / L²)

and its distance from x on negative x side = √3 L / (2√2 - √3 )

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3 years ago