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3 years ago

What is the wave speed of a wave that has a frequency of 250 Hz and a wavelength of 0.35 m?

Physics

1 answer:

azamat3 years ago

6 0

Answer:

87.5 m/s

Explanation:

The speed of a wave is given by

$v=\lambda f$

where

v is the wave speed

$\lambda$ is the wavelength

f is the frequency

In this problem, we have

$f=250 Hz$ is the frequency

$\lambda=0.35 m$ is the wavelength

Substituting into the equation, we find

$v=(0.35 m)(250 Hz)=87.5 m/s$

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castortr0y [4]

The potential energy of the block is given by:
V = m*g*h
m mass
g = 9.81m/s²
h height

The potential energy of a spring is given by:
V = 0.5 * k * x²

k spring constant
x compression of the spring

If the block starts from rest it has potential energy, but no kinetic energy. As it slides down the incline potential energy is converted into kinetic energy. When the block hits the spring the kinetic energy is converted into spring's potential energy. If the spring is fully compressed and the block is at rest again, the block has transferred all its energy into the spring. No energy is lost. So we can write:

m * g * h = 0.5 * k * x²

m = 0.5 kg
g = 9.81 m/s²
h = 2.5m * sin 37° = 1,5 m
x = 0,6 m

Solve for k.

k = 2 * m * g * h / x² = 40.8 N/m

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3 years ago

Read 2 more answers

What should be the spring constant k of a spring designed to bring a 1200 kg car to rest from a speed of 85 km/h so that the occ

borishaifa [10]

Answer:

k = 5178.8 N/m

Explanation:

As we know that spring mass system will oscillate at angular frequency given as

$\omega = \sqrt{\frac{k}{m}}$

now we have

$\omega = \sqrt{\frac{k}{1200}}$

now the maximum acceleration of the spring block system is at its maximum compression state which is given as

$a = \omega^2 A$

here A= maximum compression of the spring

so here in order to find maximum compression of the spring we will use energy conservation as we know that initial total kinetic energy of the car will convert into spring potential energy

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2$