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3 years ago

How are intensity sound and energy related

1 answer:

8 0

Sound is a form of energy in that it consists fluctuations of air pressure . The speed of the fluctuations is measured in cycles per second or Hertz (HZ)

Intensity is how large the fluctuations are, also known as amplitude and for the sound the unit is decibels of sonic pressure level (dB SPL)

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A student is bicycling north along Main Street to school. Another student is timing the bicycling student in order to determine

MatroZZZ [7]

The average velocity is -4.17 m/s

Explanation:

The average velocity of a body is given by:

$v=\frac{d}{t}$

where

d is the displacement of the body

t is the time elapsed

For the student in this problem, we have:

Initial position: $x_i = 450 m$

Final position: $x_f = 200 m$

So the displacement is

$d=x_f -x_i = 200 - 450 = -250 m$

The time elapsed is

t = 60 s

Therefore, the average velocity is

$v=\frac{-250}{60}=-4.17 m/s$

Where the negative sign means the student is moving towards the origin.

Learn more about average speed and velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

8 0

3 years ago

When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0

lora16 [44]

Answer:

$d=0.165m$

Explanation:

Given

$m=0.25kg$ , $x_{1}=5cm*\frac{1m}{100cm}=0.05m$ , $x_{2}=4cm*\frac{1m}{100cm}=0.04m$ , $v=2\frac{rev}{s}$

The tension of the spring is

$F_{k}=K*x_{1}=m*g$

$K=\frac{m*g}{x_{1}}$

$K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m$

The force in the spring is equal to centripetal force so

$F_{c}=\frac{m*v^2}{r}$

$v=w*r=2\pi*r$

But Fc is also

Fc=KxΔr

$F_{c}=K*(r-x_{2})$

Replacing

$m*4\pi^2*r=K*(r-x_{2})$

$0.25kg*4\pi^2*r=49*(r-0.04m)$

$r=0.205m$

total distance is

$d=0.205-0.04=0.165m$

3 0

3 years ago

1. Tonya had a hard time deciding between the Big Burger and the Crispy Chicken sandwiches, her two favorites.

Radda [10]

Bro the md that she lost was the md boneless burger

8 0

3 years ago

Please help! thank you

BlackZzzverrR [31]

Answer:

poor, too precise

good

Explanation:

8 0

2 years ago

A parallel-plate capacitor is disconnected from a battery, and the plates are pulled a small distance farther apart. Does Q incr

kolezko [41]

Answer:

Q stay the same

Explanation:

Charging of capacitor is done by battery . If battery is disconnected , charging will stop . There will not be any discharging as plates are separate . So pulling the plates apart will not affect the charge lying on the capacitor . It will decrease its capacity and increase its potential , keeping its charge constant.

4 0

3 years ago