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laila [671]
2 years ago
12

Which property of sound waves decreases as the square of the distance from the source increases?

Physics
2 answers:
AleksAgata [21]2 years ago
5 0
A. Pitch









okay bye have a nice day



From: Charli
Alexandra [31]2 years ago
3 0
Your answer would be A
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What is the threshold velocity vthreshold(water) (i.e., the minimum velocity) for creating Cherenkov light from a charged partic
VladimirAG [237]

Complete Question

The  complete question is shown on the first uploaded image  

Answer:

A

   v_w  =  2.256 *10^{8} \  m/s

B

  v_e  =  2.21 *10^{8} \  m/s

C

The  correct option is  B  

Explanation:

From the question we are told that

      The refractive  index of water is  n_w  =  1.33

      The  refractive  index of ethanol is  n_e  =  1.36

       

Generally the threshold velocity for creating Cherenkov light   from a charged particle as it travels through water is mathematically evaluated as

       v_w  =  \frac{c}{n_w }

Where  c is the speed of light with value  c =  3.0 *10^{8} \  m/s

       v_w  =  \frac{3.0 *10^{8}}{1.33 }

       v_w  =  2.256 *10^{8} \  m/s

Generally the threshold velocity for creating Cherenkov light   from a charged particle as it travels through water is mathematically evaluated as

            v_e  =  \frac{ c}{n_e }

  =>       v_e  =  \frac{3.0 *10^{8}}{1.36 }

=>          v_e  =  2.21 *10^{8} \  m/s

4 0
3 years ago
A 0.5 μF and a 11 μF capacitors are connected in series. Then the pair are connected in parallel with a 1.5 μF capacitor. What i
NISA [10]

Answer:

C_{eq}=1.97\ \mu F

Explanation:

Given that,

Capacitance 1, C_1=0.5\ \mu F

Capacitance 2, C_2=11\ \mu F

Capacitance 3, C_3=1.5\ \mu F

C₁ and C₂ are connected in series. Their equivalent is given by :

\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}

\dfrac{1}{C'}=\dfrac{1}{0.5}+\dfrac{1}{11}

C'=0.47\ \mu F

Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :

C_{eq}=C'+C_3

C_{eq}=0.47+1.5

C_{eq}=1.97\ \mu F

So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.

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3 years ago
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A ball is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s. How far will the ball go before hit
tia_tia [17]

The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m

s = ut + 1 / 2 at²

s = Distance

u = Initial velocity

t = Time

a = Acceleration

Vertically,

s = 15.4 m

u = 0

a = 9.8 m / s²

15.4 = 0 + ( 1 / 2 * 9.8 * t² )

t² = 3.14

t = 1.77 s

Horizontally,

u = 3.01 m / s

a = 0 ( Since there is no external force )

s = ( 3.01 * 1.77 ) + 0

s = 5.34 m

Therefore, the distance travelled by the ball before hitting the ground is 5.34 m

To know more about distance travelled

brainly.com/question/12696792

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7 0
1 year ago
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