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Nadusha1986 [10]
4 years ago
9

The sound produced by the speakers at a rock concert has a power of 0.181 W. What sound intensity do you hear when you are stand

ing 20.5 m away?
Physics
1 answer:
kiruha [24]4 years ago
8 0

Answer:

The sound intensity heard standing 20.5 m away = 0.0000343 W/m² = 3.43 × 10⁻⁵ W/m²

Explanation:

Intensity of sound at some distance r away is calculated as (power of the sound/surface area of wall of imaginary sphere at distance r)

Power of the sound = 0.181 W

Surface area of wall of imaginary sphere = 4πr² = 4π(20.5²) = 5281.02 m²

Intensity = 0.181/5281.02 = 0.0000343 W/m² = 3.43 × 10⁻⁵ W/m²

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Water flows through a garden hose which is attached to a nozzle. The water flows through hose with a speed of 1.81 m/s and throu
Serggg [28]

Answer:

a) 17.086m

b) 0.1671 m

Explanation:

Given data: speed of water through the hose  = 1.81 m/s

through the nozzle = 18.3 m/s

We know that maximum height of an object with upward velocity v is given by,

a) H = v^2/2g

where H is the maximum height water emerges  

= 18.3^2/(2×9.8) = 17.086 m answer

b) Again,  

H = v^2/2g

= 1.81^2/(2×9.8) = 0.1671 m

6 0
3 years ago
It takes 100,832 J of work to lift an elevator 18.3 meters. If this is done in 21.0 seconds, what is the average power of the el
Aneli [31]

Answer:

P=4801.5

Explanation:

Given :

work done = W = 100,832 J

time = 21.0 sec

Find:

P = ?

Formula:

P = W/t

Solution:

P = W/t

P = 100,832/21.0

  =  4801.52 J/s or Watts

6 0
4 years ago
Parallel rays from a distant object are traveling in air and then are incident on the concave end of a glass rod with a radius o
skad [1K]

Answer:

the distance of image from the vertex is 45 cm and the image formed is in the glass.

Explanation:

distance of object, u = - infinity

radius of curvature, R = - 15 cm

refractive index, n = 1.5

Let the distance of image is v.

Use the formula

-\frac{n1}{u}+\frac{n2}{v}=\frac{n2- n1}{R}\\\\-\frac{1}{\infty }+\frac{1.5}{v}=\frac{1.5-1}{-15}\\\\v=45   cm

The image is in the glass.  

8 0
3 years ago
The radius of curvature of a loop-the-loop for a roller coaster is 12.4 m. at the top of the loop (with the car inside the loop)
Ksenya-84 [330]
<span>14.79 m/s At the top of the loop, there's 2 opposing forces. The centripetal force that's attempting to push the roller coaster away and the gravitational attraction. These 2 forces are in opposite directions and their sum is 0.80 mg where m = mass and g = gravitational attraction. So let's calculate the amount of centripetal force we need. 0.80 = F - 1.00 1.80 = F So we need to have a centripetal force that's 1.8 times the local gravitational attraction which is 9.8 m/s^2. So 1.8 * 9.8 m/s^2 = 17.64 m/s^2 The formula for centripetal force is F = mv^2/r where F = force m = mass v = velocity r = radius We can eliminate mass from the equation since the same mass is being affected by both the centripetal force and gravity. So: F = v^2/r 17.64 m/s^2 = v^2/12.4 m 218.736 m^2/s^2 = v^2 14.78972616 m/s = v So the velocity at the top of the loop (rounded to 2 decimal places) is 14.79 m/s.</span>
8 0
4 years ago
A particle with charge Q and mass M has instantaneous speed uy when it is at a position where the electric potential is V. At a
olga nikolaevna [1]

Answer:

A

Explanation:

Ke = 1/2 MV^2

4 0
3 years ago
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