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3 years ago

If a speaker gives a sound intensity of 10−6w/m2 at a certain point, what is the sound intensity level β at that point?

1 answer:

5 0

Start with basic concept that you have the same amount of watts it's just how far away you are dictates the watts/meter squared (per area). As the distance from the object is increased, the area increases as distance squared (think of a sphere of increasing radius, as you increase the radius, the surface area increases as r^2, double radius, you get 4 times area, triple radius, 9 times area, and so forth). so safe from acoustical effects (reflections off walls, etc) as you increase the distance from a speaker, the sound will decrease by a factor of the distance squared, as before 2x distance =1/4 power, 3x distance =1/9 power.

so the only thing left out of your questions is what distance from speaker is the level at 10^-6 W/m^2 and what distance is point B from the speaker, and apply the 1/distance^2 principle. 

The power at point B (PB) is the following: 

power at given power level (we will call this point A or PA). 

PB=PA*(DA/DB)^2 

Where PB is power level at PB (in W/m^2) 
PA=power at point A or 10^-6 W/m^2 
DA=distance from speaker to point A 
DB=distance from speaker to point B.

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You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

5 0

3 years ago

2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her he

Ratling [72]

Answer:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, $v_s$ = d/t₂

∴ $v_s$ = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s

The minimum average speed the opponent must move so that he is in position to hit the ball, $v_s$ ≈ 5.79 m/s.

5 0

2 years ago

Which physical activity would a global positioning system resource help you with?

PolarNik [594]

The physical activity that a global positioning system resource can help me with is : ( A ) Hiking

<h3>Function of Global positioning system ( GPS ) </h3>

GPS helps with the accurate measurement of physical activities and factors such as location, time, elevation and so on. When hiking, the time, distance covered and location of the hiker can be accurately measured with the use of the GPS.

Hence we can conclude that The physical activity that a global positioning system resource can help me with is Hiking.

Learn more about GPS : brainly.com/question/9795929

#SPJ1

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1 year ago

Two people travel from Columbus to Cleveland, one by car at an average velocity of 90km/h, and one by plane at an average veloci

Flura [38]

Answer:

Explanation:

Yes , their displacement may be equal .

Suppose the displacement is AB where A is starting point and B is end point .

The car is covering the distance AB by going from A to B on straight line . On the other hand plane goes from A to C , then from C to D and then from D to B . In this way plane reaches B from A on a different path which is longer than path of the car . In the second case also displacement of plane is AB . In the second case distance covered is longer but displacement is same that is AB .

7 0

2 years ago

Why do you see objects when you shine a flashlight in a dark room?

Afina-wow [57]

Answer:

We see objects in a dark room due to the emission of light photons which are sensitive to our eyes. Darkness is simply a terminology used to describe the absence of light. Visible light to human is a component of the electromagnetic spectrum. Our eyes have receptors that picks the photons which light releases

Explanation:

3 0

2 years ago

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