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Art [367]
4 years ago
6

On a good night, the front row of the Twisted Sister concert would surely result in a 120 dB sound level. An IPod produces 100 d

B. How many IPods would be needed to produce the same intensity as the front row of the Twisted Sister concert?
Physics
1 answer:
cricket20 [7]4 years ago
8 0

Answer:

100 iPods

Explanation:

The deciBell scale is logarithmic, and thus, it turns multiplying into adding.

Initially, it was the Bell scale, purely logarithmic, where "times 10" is translated into "plus 1" (just like normal logs). However, the steps became too big and so they divided the Bell in 10 parts, the deciBell.

The levels above could well have been called 10B and 12B.

Usually, we define the dB scale for intensity as:

I(dB) = 10•log(I)

Thus,

I = 10^(I(dB)/10)

Now 120 dB gives us units of I = 10^(120/10) = 10^12 Pa (assume the dB are measured to 1 Pa) and 100 dB is 10^10 Pa.

Thus, we would need 100 ipods to get the same intensity

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A merter stick pivoted at its center has a 150gm mass suspended at its 20cm mark. a) Where should an 100gm mass be placed to pro
IgorLugansk [536]

Answer:1) 100 gm mass should be placed at 95 cm mark.

2) Mass of 112.5 gm should be placed at 90 cm mark.

Explanation:

For equilibrium of the meter stick the sum of the moment's generated by the masses should be equal and opposite

Answer to part b)

Since a meter stick is 100 cm long and it is pivoted at it's center i.e at 50 cm

Thus

1) Moment generated by 100 gm mass about center = M_{1}=m_{1}g\times r_{1}\\\\M_{1}=0.15\times 9.81\times 0.3=0.44145Nm

Let a mass 'm' be placed at 90 cm mark thus moment it generates equals

M_{2}=m\times 9.81\times 0.4=3.924m

Equating both the moments we get

0.44145=3.924m\\\\\therefore m=\frac{0.44145}{3.924}\times 1000grams\\\\\therefore m= 112.5grams

Answer to part a)

Let the 100 grams weight be placed at a distance 'x' right of center

Moment generated by 100 grams weight equals

M_{1}=0.1\times 9.81\times x

equating the moments of the forces we get

0.1\times 9.81\times x=0.15\times 0.3\times 9.81

\therefore x=\frac{0.4415}{.981}=45centimeters

thus the mass of 100 gm should be placed at 95 cm mark in the scale.

8 0
4 years ago
A small mailbag is released from a helicopter that is descending steadily at 3 m/s.
mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

       Speed of mailbag after 3 seconds = 26.4 m/s

       Package is 44.1 meter below helicopter

<u>Explanation:</u>

a)  We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

   Initial velocity = 3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

   v = 3+9.8*3 = 32.4 m/s

  Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m

  Distance traveled by package  = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

  v = -3+9.8*3 = 26.4 m/s

  Speed of mailbag after 3 seconds = 26.4 m/s

 Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m

  Distance traveled by package  = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

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