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3 years ago

The amount of work done to produce a sound determines which property of sound waves?

Physics

2 answers:

trapecia [35]3 years ago

5 0

The answer is D) Amplitude

Anna35 [415]3 years ago

4 0

Intensity of sound is defined by

$Intensity = \frac{Energy}{Area* time}$

So here we can say work done by us is stored in the form of energy of sound and that energy can be compared by the intensity of sound

So if we will do more work then the sound will have more energy and hence it will have more intensity

Now we know that intensity of sound is given as

$intensity = kA^2$

so more is the intensity then more work which means the amplitude will be more

So answer will be

D. amplitude

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Two remote control cars with masses of 1.16 kilograms and 1.98 kilograms travel toward each other at speeds of 8.64 meters per s

Black_prince [1.1K]

The initial momentum of the system can be expressed as,

$p_i=m_1u_1+m_{2_{}}u_2$

The final momentum of the system can be given as,

$p_f=m_1v_1+m_{2_{}}v_2$

According to conservation of momentum,

$p_i=p_f$

Plug in the known expressions,

$\begin{gathered} m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ m_2v_2=m_1u_1+m_2u_2-m_1v_1 \\ v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2} \end{gathered}$

Initially, the second mass move towards the first mass therefore the initial speed of second mass will be taken as negative and the recoil velocity of first mass is also taken as negative.

Plug in the known values,

$\begin{gathered} v_2=\frac{(1.16\text{ kg)(8.64 m/s)+(1.98 kg)(-3.34 m/s)-(1.16 kg)(-2.16 m/s)}}{1.98\text{ kg}} \\ =\frac{10.02\text{ kgm/s-}6.61\text{ kgm/s+}2.51\text{ kgm/s}}{1.98\text{ kg}} \\ =\frac{5.92\text{ kgm/s}}{1.98\text{ kg}} \\ \approx2.99\text{ m/s} \end{gathered}$

Thus, the final velocity of second mass is 2.99 m/s.

3 0

1 year ago

. An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?

Hoochie [10]

The acceleration of the object is $6.7 m/s^2$

Explanation:

We can solve the problem by using Newton's second law, which states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:

$F=ma$

where

F is the net force

m is the mass of the object

a is its acceleration

For the object in this problem,

F = 500 N is the applied force

m = 75 kg is the force

Solving the equation for a, we find the acceleration:

$a=\frac{F}{m}=\frac{500}{75}=6.7 m/s^2$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0

3 years ago

Which change indicates that the universe is expanding?

Aleonysh [2.5K]

Red shift of distant galaxies

5 0

3 years ago

Read 2 more answers

Two resistors, R1 and R2, are

dlinn [17]

The reciprocal of the total resistance is equal to the sum of the reciprocals of the component resistances:

1/(120.7 Ω) = 1/R₁ + 1/(221.0 Ω)

1/R₁ = 1/(120.7 Ω) - 1/(221.0 Ω)

R₁ = 1 / (1/(120.7 Ω) - 1/(221.0 Ω)) ≈ 265.9 Ω

3 0

3 years ago

Two loudspeakers in a 20c room emit 686 hz sound waves along the x-axis.a. if the speakers are in phase, what is the smallest di

jeyben [28]

Answer :

(a) d = 0.25 m

(b) d = 0.5 m

Explanation :

It is given that,

Frequency of sound waves, f = 686 Hz

Speed of sound wave at $20^0\ C$ is, v = 343 m/s

(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.

$d=\dfrac{\lambda}{2}$ ........(1)

Velocity of sound wave is given by :

$v=f\times \lambda$

$d=\dfrac{v}{2f}$

$d=\dfrac{343\ m/s}{2\times 686\ Hz}$

$d=0.25\ m$

Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.

(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.

$d=n\lambda$ ( n = integers )

Let n = 1

So, $d=\dfrac{v}{f}$

$d=\dfrac{343\ m/s}{686\ Hz}$

$d=0.5\ m$

Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.

4 0

3 years ago

Read 2 more answers