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3 years ago

The amount of energy carried by a wave is proportional to the square of the wave’s

Physics

2 answers:

nataly862011 [7]3 years ago

7 0

Answer: The correct answer is "amplitude".

Explanation:

The expression of the energy in terms of frequency is as follows;

E=hf

Here, h is Planck's constant, E is the energy and f is the frequency of the wave.

Here, the energy is directly proportional to the frequency of the wave. The energy of the wave is inversely proportional to the wavelength of the wave.

Amplitude is the maximum displacement of the particle from its equilibrium position in the vertical direction.

The amount of energy carried by a wave is directly proportional to the square of the amplitude. More the energy of the wave, more will be amplitude of the wave. Lesser the energy of the wave, lesser will be the amplitude of the wave.

Therefore, the correct option is (B).

wariber [46]3 years ago

4 0

The amount of energy carried by a wave is proportional to the square of the wave’s amplitude. The amplitude is measured by calculating the distance from the top (or bottom) of the wave and the middle or center.

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Suppose a certain jet plane creates an intensity level of 124 dB at a distance of 5.01 m. What intensity level does it create on

Zolol [24]

Answer:71 dB

Explanation:

Given

sound Level $\beta _1=124 dB$

distance $r_1=5.01 m$

From sound Intensity

$\beta =10dB\log (\frac{I_1}{I_0})$

$124=10dB\log (\frac{I_1}{I_0})$

$12.4=\log (\frac{I_1}{I_0})$

$I_1=(1\times 10^{-12})\times 10^{12.4}$

$I_1=2.51 W/m^2$

we know Intensity $I\propto ^\frac{1}{r^2}$

$I_1r_1^2=I_2r_2^2$

$I_2=I_1(\frac{r_1}{r_2})^2$

$I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2$

$I_2=1.24\times 10^{-5} W/m^2$

Sound level corresponding to $I_2$

$\beta _2=10\log (\frac{I_2}{I_0})$

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3 years ago

A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t

notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is $1.37 m/s^2$

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, $F_f$ , backward

The frictional force can be written as

$F_f = \mu mg$

where

$\mu=0.03$ is the coefficient of kinetic friction

$m=150 kg$ is the mass of the motorbike

$g=9.8 m/s^2$ is the acceleration of gravity

Therefore the net force is given by

$\sum F = F - F_f = F - \mu mg$

And substituting, we find

$\sum F=250 - (0.03)(150)(9.8)=205.9 N$

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

$\sum F = ma$