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defon
2 years ago
10

A man stands at the midpoint between two speakers that are broadcasting an amplified static hiss uniformly in all directions. Th

e speakers are 27.3 m apart and the total power of the sound coming from each speaker is 0.535 W. (Answers in W/m^2). (a) Find the total sound intensity that the man hears when he is at his initial position halfway between the speakers.
(b) Find the total sound intensity that the man hears after he has walked 4.5 m directly toward one of the speakers.
Physics
1 answer:
Illusion [34]2 years ago
5 0

Answer:

Explanation:

The power of each of the speakers is 0.535 W. At a distance d intensity of sound can be found by the following formula

Intensity of sound =  Power / 4π d²

= .535 / 4 x 3.14 x (27.3/2)²

= 2.286 x 10⁻⁴ J m⁻² s⁻¹

Intensity of sound due to other source = 5.715 x 10⁻⁵J m⁻² s⁻¹

Total intensity = 2 x 2.286 x 10⁻⁴J m⁻² s⁻¹

= 4.57 x 10⁻⁴J m⁻² s⁻¹

b ) In this case, man is standing at distances 18.15 m and 9.15 m from the sources .

The total intensity of sound reaching him is as follows

0.535 / (4 π x18.15² ) + 0.535 /  (4 π x9.15² )

= 1.293 x 10⁻⁴ + 5.087 x 10⁻⁴

= 6.38 x 10⁻⁴J m⁻² s⁻¹

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What force in Newton is required to accelerate a car starting from rest to 20 m/s in 15 seconds if the mass of the car is 2500 k
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We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.

\sf \: F=ma

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If the battery of your phone can provide 2 mA of current to your phone and holds a charge of 130 C, how long will it take a full
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Time taken for a fully charged phone to die (T) = ?

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i.e Q = I x T

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T = 130C / (2 x 10^-3A)

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Thus, it will take a fully charged phone 65000 seconds to die

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In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.
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A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is 41.9^{\circ}

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

s=u_y t + \frac{1}{2}at^2 (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

u_y = u sin \theta is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

\theta=40.0^{\circ} is the angle of projection

So

u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s

a=g=-9.8 m/s^2 is the acceleration due to gravity (downward)

Substituting the numbers, we get

-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

d=u_x t

where

u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

d=(9.19)(1.778)=16.34 m

B)

In this second case,

\theta=42.5^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s

So the equation for the vertical motion becomes

4.9t^2-8.1t-1.80=0

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s

So, the range of the shot is

d=u_x t = (8.85)(1.851)=16.38 m

C)

In this third case,

\theta=45^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s

So the equation for the vertical motion becomes

4.9t^2-8.5t-1.80=0

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s

So, the range of the shot is

d=u_x t = (8.49)(1.925)=16.34 m

D)

In this 4th case,

\theta=47.5^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s

So the equation for the vertical motion becomes

4.9t^2-8.8t-1.80=0

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s

So, the range of the shot is

d=u_x t = (8.11)(1.981)=16.07 m

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is \theta=42.5^{\circ}.

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

s-u sin \theta t + \frac{1}{2}gt^2=0

The solutions of this quadratic equation are

t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}

This is the time of flight: so, the horizontal range is

d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta

It can be found that the maximum of this function is obtained when the angle is

\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})

Therefore in this problem, the angle which leads to the maximum range is

\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0
3 years ago
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