All categories

2 years ago

Compare intensities: 80db vs 60db

1 answer:

8 0

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

$I = \frac{P}{A}$

where, P is the power through an area A.

The SI unit for I is {W/m}^{2}.

Sound intensity levels are quoted in decibels (dB) much more often than sound intensities in watts per meter squared.

The sound intensity level β in decibels of a sound having an intensity I in watts per meter squared is defined to be

$\beta (dB) = 10 log_1_0 (\frac{I}{I_o} )$

where I₀ = 10⁻¹² W/m²

80dB sound intensity is greater than 60dB sound intensity.

80dB sound intensity can be found in loud radio and classroom lecture whereas 60dB sound intensity is the normal conversation.

You might be interested in

A ball is let go and falls to the ground. It bounces a few times before

Ksenya-84 [330]

Answer:

I just want the points but you should pay attention in class jk

4 0

2 years ago

A sound wave, generated at a frequency of 440 hertz has a wavelength of 2.3 meters as it travels through a solid material. The a

Jlenok [28]

The approximate speed of the sound wave traveling through the solid material is 1012m/s.

<h3>Wavelength, Frequency and Speed</h3>

Wavelength is simply the distance over which the shapes of waves are repeated. It is the spatial period of a periodic wave.

From the wavelength, frequency and speed relation,

λ = v ÷ f

Where λ is wavelength, v is velocity/speed and f is frequency.

Given the data in the question;

Frequency of sound wave f = 440Hz = 440s⁻¹

Wavelength of the wave λ = 2.3m

Speed of the wave v = ?

To determine the approximate speed of the wave, we substitute our given values into the expression above.

λ = v ÷ f

2.3m = v ÷ 440s⁻¹

v = 2.3m × 440s⁻¹

v = 1012ms⁻¹

v = 1012m/s

Therefore, the approximate speed of the sound wave traveling through the solid material is 1012m/s.

Learn more about Speed, Frequency and Wavelength here: brainly.com/question/27120701

8 0

2 years ago

A runner begins from rest at the starting line and travles for 6.5 seconds, a runner reaches a speed of 13.4 m/s what is the run

Butoxors [25]

The acceleration of the runner in the given time is 2.06m/s².

Given the data in the question;

Since the runner begins from rest,

Initial velocity; $u = 0$

Final velocity; $v = 13.4m/s$

Time elapsed; $t = 6.5s$

Acceleration of the runner; $a = \ ?$

<h3>Velocity and Acceleration</h3>

Velocity is the speed at which an object moves in a particular direction.

Acceleration is simply the rate of change of the velocity of a particle or object with respect to time. Now, we can see the relationship from the First Equation of Motion

$v = u + at$

Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.

To determine the acceleration of the runner, we substitute our given values into the equation above.

$v = u + at\\\\13.4m/s = 0 + (a * 6.5s)\\\\13.4m/s = a * 6.5s\\\\a = \frac{13.4m/s}{6.5s}\\ \\a = 2.06m/s^2$

Therefore, the acceleration of the runner in the given time is 2.06m/s².

Learn more about Equations of Motion: brainly.com/question/18486505

3 0

2 years ago

A 10.00 kg mass is moving to the right with a velocity of 14.0 m/s. A 12.0 kg mass is moving to the left with a velocity of 8.00

Basile [38]

Answer:

2 m/s

Explanation:

From the conservation of momentum, the initial momentum of the system must be equal to the final momentum of the system.

Let the 10.00 kg mass be $m_1$ and the 12.0 kg mass be $m_2$ . When they collide and stick, they have a combined mass of $m_1+m_2$ .

Momentum is given by $p=mv$ . Set up the following equation:

$\displaystyle m_1v_1+m_2v_2=(m_1+m_2)v_f$ , where $v_f$ is the desired final velocity of the masses.

Call the right direction positive. To indicate the 12.0 kg object is travelling left, its velocity should be substitute as -8.00 m/s.

Solving yields:

$10\cdot 14 + 12\cdot (-8)=(10+12)v_f\\\implies v_f=\boxed{2 \text{ m/s}}$

4 0

2 years ago

The coefficients of friction between the load and the flatbed trailershown are μs = 0.40 and μk = 0.30. Knowing that the speed o

SOVA2 [1]

Answer:

50.97 m

Explanation:

m = Mass of truck

$\mu_s$ = Coefficient of static friction = 0.4

v = Final velocity = 0

u = Initial velocity = 72 km/h = $\dfrac{72}{3.6}=20\ \text{m/s}$

s = Displacement

Force applied

$F=ma$

Frictional force

$f=\mu_s mg$

Now these forces act opposite to each other so are equal. This is valid for the case when the load does not slide

$ma=\mu_s mg\\\Rightarrow a=\mu_s g\\\Rightarrow a=0.4\times 9.81\\\Rightarrow a=3.924\ \text{m/s}^2$

Since the obect will be decelerating the acceleration will be $-3.924\ \text{m/s}^2$

From the kinematic equations we have

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-20^2}{2\times -3.924}\\\Rightarrow s=50.97\ \text{m}$

So, the minimum distance at which the car will stop without making the load shift is 50.97 m.

5 0

2 years ago