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nata0808 [166]
2 years ago
10

What is resonance in sound​

Physics
1 answer:
Sveta_85 [38]2 years ago
6 0

Answer: Resonance in sound is when one object is vibrating at the same frequency to the second object of forces to the second frequency.

Explanation:

"Acoustic resonance is a phenomenon in which an acoustic system amplifies sound waves whose frequency matches one of its own natural frequencies of vibration (its resonance frequencies)." wikipedia I hope this helps you!

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A constant net force that has a magnitude of 135.5 N is exerted on a 26.7 kg object that is initially not moving. a. Draw a free
jek_recluse [69]

Explanation:

Given that,

Net force = 135.5 N

Mass = 26.7 kg

(a). We need to draw a free body diagram

A force is exerted on a object.

(b). We need to calculate the acceleration

Using formula of acceleration

a =\dfrac{F}{m}

a=\dfrac{135.5}{26.7}

a=5.07\ m/s^2

(c). We need to draw the sketch a position vs. time graph for the object.

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Put the value into the formula

s=0+\dfrac{1}{2}\times5.07\times t^2

s=2.535 t^2

(d). We need to draw the sketch a velocity vs. time graph for the object.

Using equation of motion

v = u+at

Put the value into the formula

v=0+5.07t

v =5.07 t

(e). We need to calculate the distance travel in 8.82 s

Using equation of motion

s=ut+\dfrac{1}{2}at^2

s=0+\dfrac{1}{2}\times5.07\times(8.82)^2

s=197.20\ m

Hence, This is the required solution.

8 0
3 years ago
Can someone help me?
Alex777 [14]

Answer:

P is pressure ,rho(slanted p) is density,g is the acceleration due to gravity, v is velocity and h is height

Explanation:

6 0
3 years ago
Ipaliwanag ang paggalang
cestrela7 [59]

paraan ito ng pagrerespeto

5 0
2 years ago
A sample of copper has a volume of 23.4 cm3 if the density of copper is 8.9 gcm3 what is the coppers mass?
murzikaleks [220]
The answer is:  " 208 g " .
_____________________________________________
Explanation:
__________________________________________
The formula/ equation for density is:
__________________________________________
D = m / V  ;  That is,  "mass divided by volume" ;
 
Density is expressed as:
__________________________________________    
                   "mass per unit volume";  in which the "mass" is expressed in units of "g" ("grams") ;  and the "unit volume" is expressed in units of:
    "cm³ " or "mL"; 
_____________________________________________
           {Note the exact equivalent:  1 cm³ = 1 mL }.
____________________________________________
         →  The formula is:  " D = m / V "  ; 
___________________________________________
   in which:

     "D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given); 

     "m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;
                 
     "V" refers to the "volume", in units of "cm³ " ;
               which is:  "23.4 cm³ " (given);
_________________________________________________
                 We want to find the mass, "m" ; so we take the original equation/formula for the density:
_________________________________________________ 
              D  =  m / V ; 
_________________________________________________________
             And we rearrange; to isolate "m" (mass) on ONE side of the    equation; and then we plug in our known/given values;
 to solve for "m" (mass);  in units of "g" (grams) ;
___________________________________________________
    Multiply each side of the equation by "V" ; 
____________________________________________________
             V * { D  =  m / V } ;  to get:
____________________________________________________
      V * D = m ;   ↔   m = V * D ;
___________________________________________________
           Now, we plug in the given values for "V" (volume) and "D" (density) ;     to solve for the mass, "m" ;
______________________________________________________
           m  =  V * D ;
 
           m  =  (23.4 cm³) * (8.9 g / 1 cm³)  = (23.4 * 8.9) g = 208.26 g ;
  
 →  Round to "208 g" (3 significant figures);  
____________________________________
The answer is:  " 208 g " .
_____________________________________________________
7 0
3 years ago
With certain exceptions, Class E airspace extends upward from either 700 feet or 1,200 feet AGL to, but does not include,A) 14,5
DedPeter [7]

Answer:

B) 18,000 feet MSL

Explanation:

There are three-dimensional parts in the navigation airspace in the world. The class E airspace is mostly used in the regions with coastal areas that are relatively populated. If we consider certain forms of exceptions, the class E airspace can move in the upward direction to few feet (i.e. 1200 ft). However, this doesn't include 18,000 feet MSL.

6 0
3 years ago
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