The Wavelength and pressure amplitude of the final note is 0.39 m and 0.255 Pa respectively.
What do you mean by the pressure amplitude?The minuscule pressure difference between compressions and rarefactions in a sound wave is what the human ear uses to function. Pressure amplitude refers to the size of these pressure variations. Generally speaking, a sound appears louder the larger the pressure amplitude it has. Even for very loud sounds, these pressure changes are negligible in comparison to ambient pressure. A jet plane at 30 yards, for example, has pressure changes of just a few hundred Pascals, or less than one-hundredth of atmospheric pressure, which is loud enough to cause damage to human ears (which is roughly 100,000 Pa). The pressure amplitudes of the quietest noises that people can hear are approximately one billionth of the atmospheric pressure.
x=Asin (w [ t-K])+ b
Given:
Initial frequency, [tex]f_{initial} = 146.8\;Hz[/tex]
Final frequency, [tex]f_{final} = 880\;Hz[/tex]
Sound level, [tex]\beta=79\;dB[/tex]
We know that,
Speed of the sound, [tex]v=343\;m/s[/tex]
Density of air, [tex]\rho=1.2\;kg/m^{3}[/tex]
Wavelength, [tex]\lambda=\frac{v}{f}[/tex]
Wavelength of final note is,
[tex]\lambda=\frac{v}{f_{final}}[/tex]
Substitute the known values in the above equation,
[tex]\lambda=\frac{343}{880}}[/tex]
[tex]\lambda=0.39\;m[/tex]
[tex]\beta=10\times log(\frac{I}{I_{0}} )[/tex]
We know that, [tex]I_{o}=1 \times 10^{-12}\;W/m^{2}[/tex]
[tex]79=10\times log(\frac{I}{1 \times 10^{-12}} )[/tex]
By solving the above equation, We get,
[tex]I=7.9\times 10^{-5} W/m^{2}[/tex]
Pressure amplitude of the final note is,
[tex]I = \frac{(\triangle p)^{2} }{2 \times \rho \times v}[/tex]
[tex](\triangle p)^{2} = I \times2 \times \rho \times v[/tex]
[tex]\triangle p = \sqrt{I \times2 \times \rho \times v}[/tex]
Substitute the known values in the above equation,
[tex]\triangle p = \sqrt{7.9 \times 10^{-5} \times2 \times 1.2 \times 343}[/tex]
By solving the above equation, We get,
[tex]\triangle p = 0.255\;Pa[/tex]
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what's newtons second law
Newton's Second Law of Motion states "The rate of change of momentum of a body is directly proportional to the applied unbalanced force in the direction of force".
An nfl caliber wide receiver can reach a top speed of 10.0 m/s in only 2.4 s. what is their acceleration during this time?
The nfl caliber wide receiver that can reach a top speed of 10.0 m/s in only 2.4 s has an acceleration during this time of: 4.16 m/s²
The formula and procedure we will use to solve this exercise is:
a = (vf - vi) /t
Where:
a = accelerationvf = final velocityvi = initial velocityt = timeInformation about the problem:
vi= 0 m/svf = 10.0 m/st = 2.4 sa=?Applying the acceleration formula we have:
a = (vf - vi) /t
a = (10.0 m/s - 0 m/s) /2.4 s
a = (10.0 m/s) /2.4 s
a = 4.16 m/s²
What is acceleration?It is a physical quantity that indicates the variation of velocity as a function of time, it is expressed in units of distance per time squared e.g.: m/sec2 ; km/h2
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A projectile has an initial horizontal velocity of 34.0 M/s at the edge of a roof top. Find the horizontal and vertical components of its velocity after 5.5s
I need shown work !
Answer:
[tex]v_x=34 m/s[/tex]
[tex]v_y=53.9\ m/s[/tex]
Explanation:
Horizontal Launch
When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.
The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:
vx=v
The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:
[tex]v_y=g.t[/tex]
The horizontal component of the velocity is always the same:
[tex]v_x=34 m/s[/tex]
The vertical component at t=5.5 s is:
[tex]v_y=9.8*5.5=53.9[/tex]
[tex]v_y=53.9\ m/s[/tex]
Suppose our Sun is about to explode. In an effort to escape, we depart in a spacecraft at v=0.800 c and head toward the star Tau Ceti, 12.0 ly away. When we reach the midpoint of our journey from the Earth, we see our Sun explode, and, unfortunately, at the same instant, we see Tau Ceti explode as well.(b) What If? In a frame of reference in which the Sun and Tau Ceti are at rest, did they explode simultaneously? If not, which exploded first?
Considering a scenario in which the Sun is going to explode.
In order to escape the explosion, we depart in a spacecraft with a speed of v = 0.8c.
The star Tau Ceti is 12 life years away.
At the midpoint of the journey, the Sun as well as Tau Cetik explode. at the same instant.
Imagine a hermit who is immobile with respect to both the Sun and Tau Ceti and resides on an asteroid that is midway between both. He observes the blast waves of both explosions as our spaceship passes him. This observer concludes that the two stars blew up simultaneously because he believes both stars to remain stationary.
In the frame of reference, the Sun and Tau Ceti explode at the same time and will remain stationary.
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Descibe the real-world examples of Newton's third lawthat were idenified in "Applications of Newton's Laws."
Answer:
A horse pulls a cart, a person walks on the ground
Explanation:
Answer:
Know it's late but I couldn't find a good answer when I looked for it so i did it myself.
My answer:
One of the real world examples mentioned in the article was that of a rocket blasting off into space. It was explained that it's an example of Newton's third law because the rocket’s engines push on those gases (action force), and the gases push back (reaction force). This reaction shows an excellent example of Newton's third law.
Sample Response:
When gases escape downward out the tail of a rocket, they push upward on the rocket. When layers of the Earth collide, they force the Earth upward, forming mountains.
Explanation:
hope this helps <3
Explain how energy from the sun is the primary source of energy on planet Earth, using the principles of the conservation of energy and examples to support your answer.
PLEASE HURRY
Which of the following is a contact force? a. magnetic
b. applied
c. gravity
d. electric
Answer:
B. Applied
Explanation:
Applied means that somethings is pushing directly into it like a person to a box. and well the reasons why the others are not contact force are because gravity its cause to an attraction to the planet and no matter where you are it will still attract you, electric is not because it can flow through air, magnetic it affects to an invisible field that even if its not torching the marital you can still feel the magnetic push or pull
So applied
a student is measuring the effects of an electric circuit on a compass, and he is testing to see if increasing the current will cause the compass needle to change its direction faster. he does one trial using a 9-volt battery and another trial using a 1.5-volt battery. he takes accurate data. based on good experimental design, what would improve his experiment? he should add a non working compass to act as a control. he should add a non working compass to act as a control. he should use batteries of the same voltage but from different manufacturers to show he has used different variables. he should use batteries of the same voltage but from different manufacturers to show he has used different variables. he should repeat the test with three additional batteries of varying voltages to demonstrate the reproducibility of results. he should repeat the test with three additional batteries of varying voltages to demonstrate the reproducibility of results. he should use only one battery but do several trials, using different circuit designs to ensure he has appropriate test groups. he should use only one battery but do several trials, using different circuit designs to ensure he has appropriate test groups.
He should repeat the test with three additional batteries of varying voltages to demonstrate the reproducibility of results. he should use only one battery but do several trials, using different circuit designs to ensure he has appropriate test groups - this can improve the student's experiment.
What is reproducibility?Reproducibility, otherwise called replicability and repeatability, is a significant rule supporting the logical technique. For the discoveries of a review to be reproducible implies that results got by an examination or an observational review or in a factual investigation of an informational index. Albeit numerous organic researchers instinctively accept that the reproducibility of a trial implies that it tends to be repeated, Drummond makes a differentiation between these two terms. Informational indexes are huge and calculations are more refined. This requests reproducibility to decrease the mistake and predisposition when people are added to the course of information investigation.
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NEED ANSWER ASAP
Study the circled features on the surface of the Sun.
The circled areas are called_____. They are_______ the surrounding area.
Answer:
the circles are called Sun spots. they are colder than the surrounding area.
Explanation:
Answer:
The circled areas are called__Sunspots___. They are___colder____ the surrounding area.
Explanation:
i just took the test it is correst plz give me a thanks or a brainiest and ill give 1 back :D
In an electrical circuit, the term current refers to:
a. resistance.
b. potential difference.
c. flowing charges.
d. energy loss.
Answer: c
Explanation:
Current is the flowing charges that goes around p.d increases it.
a helicopter is ascending at a rate of 14m/s at a height of 20m above the ground it drops a package Calculate the time taken by the package to reach the ground
Adding the rise time and fall time then the package to reach the ground in 3.647 second.
Time can be defined in the sence as,that it is a continued sequence of existence and events that occurs in an apparently irreversible succession from the past, through the present, into the future.
We are given that ,
The packege 's trajectory into a rise to maximum height and fall from there to the ground .Rise to maximum height begins at vertical position is given as,
Vertical position of the helicopter = H₀ = 20 m,
Initial velocityof helicopter = u = 14 m/s upward
Covered time of the helicopter =t₁ = 0,
Acceleration under gravity of helicoper=(- a) = -9.81 m/s²
Rise ends when instantaneous velocity v = 0,
Therefore, from the equation of motion for velocity as a function of time is written as ,
v = a + at
0 = 14 m/s - (9.81 )m/s² t₁
t₁ = 1.427sec
Thus , the velocity changes linearly over time t₁ sec from u = 0 m/s then the averaging velocity can be given by w i.e.
w = u /2 = (14m/s)/2
w = 7 m/s
Therefore, maximum height of packege is given as,
H₁ = H₀ + w × t₁
H₁ =20m + 9.81m/s (1.427)s
H₁ = 33. 998 m
Therefore, when the packege fall from maximum height begins at ,
Time= t₁ = 0,
Position= h = H₁
Velocity= u = 0, and
Acceleration due to gravity downward=(-a) = -9.81 m/s²
Thus the equation of motion for position (s) as a function of initial position (h) is can be written as,
s = h + ut₁ + (1/2) at₂²
0 = H₁ + 0 + (1/2) (-9.81)t₂²
t₂² = 4. 890 s
t₂ = 2.21 s
Hence , on adding the rise time t₁ and fall time t₂
Total time T = t₁ + t₂
T = 1.427 s + 2.21s = 3.647 s
Thus, then the package to reach the ground in 3.647 second.
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Which SI units are combined to describe a force
A. Kilograms and m/s
B. Kilograms and meters
C. Kilograms and m/s^2
D. Kilograms and joules
Answer:
C.
Explanation:
kilograms and m/s^2
Question 2 (1 point)
If you drove at a speed of 20 km/h for 2 hours, what distance did you cover?
а
b
10 km
40 km
22 km
с
d
20 km
Question 311 point)
Answer:
ewwwwwwwwwwww
get it out!
two identical traveling waves, moving in the same direction, are out of phase by π/2 rad. what is the amplitude of the resultant wave in terms of the common amplitude ym of the two combining waves?
The amplitude of the resultant wave in terms of the common amplitude [tex]y_m[/tex] of the two combining waves which are out of phase by [tex]\frac{\pi}{2} \ \text{rad}[/tex] is [tex]\sqrt{2} y_m[/tex].
The amplitude of two waves interfering with each other depends on the principle of superposition.
Since, the waves are out of phase by [tex]\frac{\pi}{2} \ \text{rad}[/tex] so we can write the waves to be:
[tex]y_1=Asin(kx-\omega t)[/tex]
[tex]y_2=Asin(kx-\omega t+\frac{\pi}{2})\\ \therefore y_2=Acos(kx-\omega t)[/tex]
According to the principle of superposition, the resultant wave is written as:
[tex]y=y_1+y_2[/tex]
[tex]\Rightarrow y=A[sin(kx-\omega t)+cos(kx-\omega t)]\\\Rightarrow y=\sqrt{2}A[\frac{1}{\sqrt{2}}sin(kx-\omega t)+\frac{1}{\sqrt{2}}cos(kx-\omega t)]\\y=\sqrt{2}Asin(kx-\omega t+ \frac{\pi}{4})[/tex]
So from the above relation, we have the amplitude to be:
New Amplitude = [tex]\sqrt{2}A[/tex]
Hence, the amplitude of the combined waves is [tex]\sqrt{2}A[/tex].
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What are brown bears doing as a result of panting?
maintaining internal water levels by releasing moisture
maintaining internal water levels by taking in moisture
maintaining body temperature by warming up
maintaining body temperature by cooling down
They are maintaining their body temperature by cooling down, like dogs do, as they don't have enough sweat glands to keep them cool without panting.
Brown bears maintain body temperature by cooling as a result of panting. So, the correct option is D.
What is Body temperature?Normal human body-temperature is defined as the typical temperature range found in humans which is commonly referred to as 36.5–37.5 °C. The human body temperature varies. Temperature is known as the degree of hotness or coldness of an area or object.
When we are very hot, the blood vessels of the skin dilate to carry the excess heat to the surface of the skin. Brown bear panting ensures that cool air circulates over the moist surfaces in the mouth which helps maintain body temperature by cooling the body.
Therefore, the correct option is D.
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A doctor is examining a child with a red, inflamed ear
canal. The child is suffering some hearing loss. The ear,
contains a small amount of soft earwax.
Which is most likely the cause of the child's hearing
loss?
O damage to the outer ear
O a disease or infection of the outer ear
O damage to the bones of the middle ear
O a disease or infection of the inner ear
Answer:
C
Explanation:
True or False. Water drains differently in the Northern and Southern hemispheres.
Answer: True
Explanation:
Answer:
false
Explanation:
hope this help (brainlist plz)
Two parallel wires are separated by 6.00 cm, each carrying 3.00 A of current in the same direction.(b) Is the force attractive or repulsive?
The force is attractive.
The force between two parallel currents I₁ and I₂ that are spaced apart by r is measured as the ratio F/L. If the currents are flowing in the same direction, the force is attracting; if not, it is repulsive.
Each parallel wire carries 3.00 A of current in the same direction and is spaced apart by 6.00 cm.
Think of these two wires as being placed side by side on a table, with currents flowing in opposite directions toward you—wire 1 to the left and wire 2 to the right. The magnetic field produced by wire 1 at the location of wire 2 is directed vertically upward, according to the right-hand rule that links current to field direction. The force experienced by wire 2 as a result of its current flowing through this field is therefore determined by the right-hand rule, which links current and field to force, to be to the left, back toward wire 1. The force that one wire applies to another is therefore an attractive force.
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What is the acceleration of a 7 kg mass being pulled by a 56 N force?
Answer:
8 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
m is the mass
f is the force
From the question we have
[tex]a = \frac{56}{7} \\ [/tex]
We have the final answer as
8 m/s²Hope this helps you
Ricardo came up with three different designs for paper airplanes. He would like to find out which design flies the farthest. Ricardo plans to fly each paper airplane five times. What step should ricardo take after he flies each plane?.
The three different paper airplane designs were created by Ricardo. In an effort to determine which design flies the farthest, Ricardo intends to fly each paper airplane five times. Following each flight, Ricardo should measure the distance flown by the aircraft and note it.
How do you gauge a distance?Two of the most frequent techniques used in agricultural surveying involve measuring angles and distances with basic tools. even when the tools are basic. With diligence and practice, a great degree of precision can be attained.
The (1) pace is the main technique for gauging distance. 2) The odometer. (3) chaining or taping. four stages. optical range finder, number 5. and (6) techniques for electronic distance measurement (EDM).
It will be beneficial to review the standard units of distance first (displacement). The following are the standard English units:
12 inches make up one foot (in)
Yd (yard) = 3 feet (ft)
1 rod is equal to 16.5 feet and 5.5 yards (yd)
A mile is equal to 5280 feet, 1760 yards, and 320 rods.
When converting between different units of measurement, the units cancellation approach comes in quite handy.
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6. A bored student throws a pebble straight up from a 5m cliff at 12 m/s. How much time
does it take to hit the ground below? Part 2: which letter (A, B or C) most accurately
portrays this scenario?
Answer:
It takes 0.416 seconds to hit the ground.
Explanation:
speed = distance/ time
12 = 5/ x (multiply both with x)
12x = 5 (÷ 12)
x = 0.416 s
what are the two components of any vector
Answer:
U^U
Explanation:
Two-dimensional vectors have two components: an x vector and a y vector. Each of these vector components is a vector in the direction of one axis. The sum of the components of vectors is the original vector. Three-dimensional vectors have a z component as well.
A cannonball with a perfect horizontal velocity of 80.0 m/s flies 320 m before hitting the ground. How much time was the cannonball in the air?
Time taken to reach height is 4 seconds.
Given that;
Velocity of cannon ball = 80 m/s
Height taken = 320 meter
Find:
Time taken to reach height
Computation:
S = ut + (1/2)gt²
320 = (80)(t) + (1/2)(10)(t²)
320 = 80t + 5t²
t² + 16t + 64
So,
t = 4
Time taken = 4 seconds
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write an essay for me pls about Proxima Centauri
this will help a lot
Answer:
Proxima Centauri is a small, low-mass star located 4.244 light-years (1.301 pc) away from the Sun in the southern constellation of Centaurus. Its Latin name means the "nearest [star] of Centaurus". This object was discovered in 1915 by Robert Innes and is the nearest-known star to the Sun.
Star system: Alpha Centauri
Category: Star, Flare star
Constellation: Centaurus
Discoverer: Robert T. A. Innes
Explanation:
u should make it into an essay
the following measurements are ranked from largest to smallest: 11.6 mg, 1021 µg, 0.000006 kg, 0.31 mg.
The ranking of the measurement from the largest to the smallest is 11.6 mg > 0.000006 kg > 1021 µg > 0.31 mg.
What is ranking of measurement?Ranking of measurement is a classification that describes the nature of information within the values assigned to variables.
This is usually done based on the size of each item in the list of measurement.
11.6 mg = 0.0116 g
1021 µg = 0.001021 g
0.000006 kg = 0.006 g
0.31 mg = 0.00031 g
Thus, the ranking of the measurement from the largest to the smallest is 11.6 mg > 0.000006 kg > 1021 µg > 0.31 mg.
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how many years would it take to reach the star from earth, as measured by observers on the spacecraft
In other words, it would take Deep Space 1 more than 81,000 years to travel the 4.24 light-years between Earth and Proxima Centauri at its top speed of 56,000 km/h. In relation to human history, that would be more than 2,700 generations.
Nearly 40 trillion kilometers, or 4.4 light-years, separate us from Alpha Centauri. The NASA-Germany Helios probes, the fastest spacecraft to date to be launched into orbit, flew at a speed of 250,000 kilometers per hour. The probes would need 18,000 years to travel at such pace to arrive at the sun's nearest neighbor. The calculations reveal that it is almost impossible to reach the nearest star in a human lifetime, even with the most futuristic technologies.
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A projectile was launched horizontally with a velocity of 468 m/s, 1.86 m above the ground. Calculate how long it would take for the projectile to hit the ground, to the nearest tenth of a second. Record your answer in the boxes below. Be sure to use the correct place value.
Answer:
0.6 seconds
Explanation:
The time to fall from height h is ...
t = √(2h/g)
t = √(2(1.86 m)/(9.8 m/s^2)) ≈ √0.3796 s ≈ 0.616 s
It would take about 0.6 seconds for the projectile to hit the ground.
Answer:
0.6 seconds
is ur answer :)
✌❤
A 1.3 kg mass starts from rest and slides down an inclined plane 8.7 ✕ 10-1 m long in 0.37 s. What net force is acting on the mass along the incline?
Answer:
The net force acting on the mass is 16.523 N
Explanation:
The given mass of the object, m = 1.3 kg
The length of the inclined plane over which the mass slides = 8.7 × 10⁻¹ m
The time it takes the mass to slide = 0.37 s
Therefore, we apply the following kinematic equation of motion;
s = u·t + 1/2·a·t²
Where;
u = The initial velocity of the mass = 0 m/s
t = The time taken for the motion = 0.37 s
a = The acceleration
s = The distance moved in the motion = 8.7 × 10⁻¹ m = 0.87 m
Plugging in the values gives;
0.87 = 0 × t + 1/2 × a × 0.37²
∴ 0.87 = 1/2 × a × 0.37²
a = 0.87/(1/2 × 0.37²) ≈ 12.71 m/s²
The net force acting on the mass is F = Mass, m × Acceleration, a force = m × a
Therefore, the net force acting on the mass is F = 1.3 × 12.71 = 16.523
The net force acting on the mass, F = 16.523 N.
The net force acting on the mass along the incline is 16.523 N
From the information given:
the mass of the object = 1.3 kgthe length which corresponds to the distance (S) of the inclined plane = 8.7 × 10⁻¹ m the time taken = 0.37 sAccording to the second equation of motion;
[tex]\mathbf{S = ut + \dfrac{1}{2}at^2}[/tex]
the initial velocity u = 0 m/s since the mass starts from the rest
∴
[tex]\mathbf{S =0(t) + \dfrac{1}{2}at^2}[/tex]
[tex]\mathbf{S = \dfrac{1}{2}at^2}[/tex]
2S = at²
[tex]\mathbf{a = \dfrac{2S}{t^2}}[/tex]
[tex]\mathbf{a = \dfrac{2\times 0.87 \ m}{0.37^2}}[/tex]
acceleration (a) = 12.71 m/s²
The net force now acting on the mass along the incline can now be estimated as:
[tex]\mathbf{F_{net} = ma}[/tex]
[tex]\mathbf{F_{net} = 1.3 \ kg \times 12.71 \ m/s^2}[/tex]
[tex]\mathbf{F_{net} = 16.523 \ N}[/tex]
Therefore, we can conclude that the net force acting on the mass along the incline is 16.523 N
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A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the magnitude of the electric field(d) 60.0 cm from the center of the sphere.
The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
[tex]Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}[/tex]
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
[tex]E=\dfrac{Q}{4\pi\epsilon_0 r^2}[/tex]
Substitute numerical values:
[tex]E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}[/tex]
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
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A 50.5 N object is in free fall.
What is the magnitude of the net force
which acts on the object?
Answer:
-49.49
Explanation:
If there’s no air resistance then the only thing acting on the object is the force of gravity which is mass times gravity. And to get mass all we have to do is divide 50.5/10 then times that by -9.8 since that’s the force of gravity.
Answer: The answer would be 50.5 N.
Explanation: It’s because there’s no air resistance.