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melamori03 [73]
3 years ago
13

Describe how noise-cancelling headphones modify the sound waves around the user. Your answer should be three to four sentences i

n length and contain proper grammar and punctuation.
Physics
1 answer:
amm18123 years ago
3 0
<h3><u><em>Answer: Describe how noise-cancelling headphones modify the sound waves around the user. Your answer should be three to four sentences in length and contain proper grammar and punctuation?  How Noise-canceling Headphones Work </em></u></h3><h3><u><em> </em></u></h3><h3><u><em> </em></u></h3><h3><u><em>Noise-canceling Headphones </em></u></h3><h3><u><em> </em></u></h3><h3><u><em>Bose was the first company to introduce noise-canceling headphones. </em></u></h3><h3><u><em> </em></u></h3><h3><u><em>Bose was the first company to introduce noise-canceling headphones. </em></u></h3><h3><u><em> </em></u></h3><h3><u><em>PHOTO COURTESY CONSUMER PRODUCTS </em></u></h3><h3><u><em> </em></u></h3><h3><u><em>Unfortunately for music lovers, many types of ambient sounds can interfere with or even block the sounds coming through their headphones. If you have ever tried to listen to a CD or MP3 player on a plane, then you know the problem well: The roar of the engines makes it difficult to hear what's being piped through the speakers -- even when those speakers are situated in or on your ear. Fortunately, noise-canceling headphones can provide a more enjoyable listening experience. </em></u></h3><h3><u><em> </em></u></h3><h3><u><em>Noise-canceling headphones come in either active or passive types. Technically speaking, any type of headphones can provide some passive noise reduction. That's because the materials of the headphones themselves block out some sound waves, especially those at higher frequencies. The best passive noise-canceling headphones, however, are circum-aural types that are specially constructed to maximize noise-filtering properties. That means they are packed with layers of high-density foam or other sound-absorbing material, which makes them heavier than normal headphones. The tradeoff of all that extra weight is a reduction in noise of about 15 to 20 decibels (dB). But considering jet engines create 75 to 80 dB of noise inside the aircraft cabin, passive models have some serious limitations. That's where active noise-canceling headphones come in. </em></u></h3><h2><u><em> </em></u></h2>

<u><em></em></u>

<h2><u><em>Hope this Helps you!! Good Luck with your Quiz/Assignment/Test!! Don't forget to mark as Brainliest!! :)</em></u></h2>

<u><em /></u>

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A disk-shaped merry-go-round of radius 2.93 m and mass 165 kg rotates freely with an angular speed of 0.691 rev/s . A 62.4 kg pe
vekshin1

Answer:

Explanation:

The problem is related to rotational motion . So we shall find out rotational kinetic energy .

K E = 1/2 x I ω²

ω is the final angular velocity

Moment of inertial of the disk

I ₁ = 1/2 m r²

= .5 x 165 x 2.93²

= 708.25 kgm²

Moment of inertial of the person

I₂ = mr²

= 62.5 x 2.93²

= 536.55 kgm²

ω₂ = v / R

= 3.11 / 2.93 rad /s

At the time of jumping , law of conservation of angular momentum will apply

I₁ ω₁ + I₂ω₂ =  (I₁  + I₂)ω

708.25 x0.691 + 536.55 x ( 3.11 / 2.93 ) = ( 708.25 + 536.55 ) ω

ω = 0  .85 rad/ s

K E = 1/2 x I ω²

= .5 x ( 708.25 + 536.55 ) ( .85 )²

449.68 J

3 0
3 years ago
Read 2 more answers
PLZ HELP ON #22-26!!!! <br><br>Please explain why and how you got your answer.
AleksAgata [21]
22. a - (vf^2 - vi^2)/(2d) 
a = (0 - 23^2)/(170) 
a = -3.1 m/s^2

23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3 
33 = 3t 
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long. 

24. The sprinter starts from rest. The average acceleration is found from: 
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s

25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m 

26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s 
acceleration, a = -7.00m/s^2
displacement, s - 92m 
Using v^2 = u^2 - 2as 
0^2 - u^2 + 2 (-7.00) (92) 
initial velocity, u = sqrt (1288) = 35.9 m/s 
This is the speed pf the car just bore braking. 

I hope this helps!! 

5 0
3 years ago
The function H(t) = -16t^2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the g
Reptile [31]
<span>H(t) = -16t^2 + vt + s
</span><span>Part A:
</span>Using the given data:

H(t)= -16*t² + 60*t + 82;

Part B:
Put H(t)=0
0<span>= -16*t² + 60*t + 82;</span>
Use the quadratic formula to find t.
See the attachment...'t' is replaced with 'x'.

3 0
2 years ago
A common flashlight bulb is rated at 0.23 a and 2.9 v (the values of the current and voltage under operating conditions). if the
allochka39001 [22]
The resistance at operating temperature is R = V/I = 2.9 V / 0.23A = 12.61 ohmsT from R – R0 = Roalpha (T – T0), we find that:T = T0 + 1/alpha (R/R0 -1) = 20 degrees Celsius + (1/ 4.3 x 10^-3/K) (12.61 ohms/ 1.1 ohms – 1)T = 2453.40 degrees Celsius
6 0
3 years ago
The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in
maria [59]

Answer: (a). Resistance = 0.4286ohms and Current (I) = 7A

(b). Resistance (R) = 0.027 ohms and Current (I) = 111.1A

(c). Resistance (R) = 0.1071 ohms and Current (I) = 28A

Explanation:

From the question, given that;

ρ = 1.5*10-2ῼ.m

Lo = 7cm = 0.07m

V = 3V

From the formula R = ρL/A, where A is the area of cross section, L is the length of material and ρ is the resistivity.

(A)

L = 4Lo and A = 2Lo*Lo

R = ρL/A

R = ρ4Lo/(2Lo*Lo)

R = 2ρ /Lo = 2*1.5*10-2/0.07

R = 0.4286 ῼ

From this the current becomes;

I = V/R = 3/ 0.4286 = 6.99 = 7A

(B)

L = Lo and A = 4Lo * 2Lo

R = ρL/A

R = ρLo/ (4Lo*2Lo) after eliminating Lo from both sides we get,

R = ρ/8Lo = 1.5*10-2 / 8*0.07

R = 0.027

Current (I) = V/R = 3/0.027 = 111.1A

(C)

L = 2Lo and A = Lo * 4Lo

R = ρL/A

R = ρ2Lo/ (Lo*4Lo) eliminating Lo from both sides we get,

R = ρ/2Lo = 1.5*10-2 / 2*0.07 = 0.1071

The current becomes;

I = V/R = 3/0.1071 = 28A

4 0
3 years ago
Read 2 more answers
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