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Naily [24]
2 years ago
6

Sound is an example of a ____ Wave.

Physics
2 answers:
sesenic [268]2 years ago
7 0

Answer:

Mechanical Waves

Explanation:Sound waves in air are longitudinal waves because particles of the medium through which the sound is transported vibrate parallel to the direction that the sound wave moves. A vibrating string can create longitudinal waves.

olga2289 [7]2 years ago
3 0

Answer:

Mechanical Wave

Explanation:

Sound is an example of a Mechanical Wave.

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In the decomposition of water, why is twice as much hydrogen as oxygen formed?
astra-53 [7]

Answer:

I think

D : Hydrogen gas has two atom , and oxygen has one atom

7 0
3 years ago
If you wanted to know the location of a vehicle that ran out of gas after taking a zigzag route through the city, which quantity
Papessa [141]

Explanation:

the vehicles displacement, since displacement deals with position

8 0
3 years ago
A child on a swing sweeps out a distance of 45 ft on the first pass. If she is allowed to continue swinging until she​ stops, an
earnstyle [38]

Answer:

d = 90 ft

Explanation:

Here in each swing the distance sweeps by the swing is half of the initial distance that it will move

So here we can say that total distance in whole motion is  given as

d = 45 + \frac{45}{2} + \frac{45}{4} + \frac{45}{8}..........

since it is half of the distance that it will move in each swing so it would be a geometric progression with common ratio of 1/2

so sum of such GP is given by the formula

S = \frac{a}{1 - r}

d = \frac{45}{1 - \frac{1}{2}}

d = 90 ft

6 0
3 years ago
you read a primary source and a secondary source that discuss the same experiment. There is a difference in the conclusion made
Nuetrik [128]
You should trust the primary source more.

This is because the primary source is make its conclusion from direct observation, while the secondary source is possibly making reference to another secondary source or to another primary.

The primary source should be trusted more because it is from direct observation.     
4 0
3 years ago
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value v_2  =  4 \sqrt{10} \  m/s

Explanation:

From the question we are told that

   The  charge on the first sphere is  q_1  =  2\mu C  =  2*10^{-6} \  C

    The charge on the second sphere is  q_2 =  8 \mu C = 8*10^{-6} \  C

     The  mass of the second charge is m  =  1.50 \  g  =  1.50 *10^{-3} \ kg

      The  distance apart is  d =  0.4 \  m

      The  speed of the second  sphere is  v_1  =  20 \  ms^{-1}

Generally the total energy possessed by when q_2 and  q_1 are separated by 0.8 \  m is mathematically represented

     Q =  KE + U

Here KE   is  the kinetic energy which is mathematically represented as

     KE  =  \frac{1 }{2}  m (v_1)^2

substituting value

     KE  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (20 )^2

     KE  =  0.3 \  J

And  U is  the  potential  energy which is mathematically represented as

        U  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.8 }

      U  =  0.18 \  J

So

       Q =  0.3 +  0.18

       Q =  0.48 \  J

Generally the total energy possessed by when q_2 and  q_1 are separated by 0.4 \  m is mathematically represented

         Q_f =  KE_f + U_f

Here KE_f is  the kinetic energy which is mathematically represented as

     KE_f  =  \frac{1 }{2}  m (v_2^2

substituting value

     KE_f  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (v_2 )^2

     KE_f  =  7.50 *10^{ -4} (v_2 )^2

And  U_f is  the  potential  energy which is mathematically represented as

        U_f  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U_f  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.4 }

      U_f  =  0.36 \  J

From the law of energy conservation

     Q =  Q_f

So

    0.48 =  0.36 +(7.50 *10^{-4} v_2^2)

   v_2  =  4 \sqrt{10} \  m/s

     

   

6 0
3 years ago
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