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ohaa [14]
3 years ago
13

Two identical loudspeakers are driven in phase by a common oscillator at 750 Hz and face each other at a distance of 1.24 m. Loc

ate the points along the line joining the two speakers where relative minima of sound pressure amplitude would be expected. (Take the speed of sound in air to be 343 m/s. Choose one speaker as the origin and give your answers in order of increasing distance from this speaker. Enter 'none' in all unused answer boxes.)
Physics
1 answer:
Juli2301 [7.4K]3 years ago
6 0

Answer:

0.2286  m,  0.686 m and  1,143 m

therefore we see that there is respect even where the intensity is minimal

Explanation:

Destructive interference to the two speakers is described by the expression

       Δr = (2n +1) λ/2

where r is the distance,  λ the wavelength and n an integer indicating the order of the interference

let's locate the origin on the left speaker

let's find the wavelength with the equation

        v =  λ f

         λ = v / f

we substitute

      Δr = (2n + 1) v / 2f

let's calculate for difference values ​​of n

     Δr = (2n +1) 343/(2 750)

     Δr = (2n + 1) 0.2286

we locate the different values ​​for a minimum of interim

    n     Δr (m)

     0    0.2286

     1     0.686

     2    1,143

therefore we see that there is respect even where the intensity is minimal

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faust18 [17]

The forces acting on the elevator are:

Gravity force

Tension force

Air resistance

Explanation:

Let's go through each of the forces listed and see which ones are acting on the elevator.

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  • Gravity force: YES. The force of gravity acts on every object located in the gravitational field of the Earth; it pulls downward, and its magnitude is mg, where m is the mass of the object and g is the acceleration of gravity.
  • Applied force: NO. Here there is no applied force, since there is nobody "pushing" or "pulling" the elevator.
  • Friction force: NO. As we are considering the forces on the elevator, and the elevator is not sliding against any surfaces, there is no force of friction. (The force of friction acts whenever there are two surfaces sliding against each other, which is not the case here)
  • Tension force: YES. The tension force is the force exerted by a rope or a string when pulling an object. In this case, there are four ropes pulling the elevator, therefore there are 4 forces of tension acting on the elevator, upward.
  • Air resistance: YES. As the elevator is moving through the air, the interaction between the molecules of air with the surface of the elevator produces a force (called air resistance) that "resists" the motion of the elevator, therefore pushing downward. However, the magnitude of this force is negligible in this case.

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A motorcycle begins at rest and accelerates uniformly S7.9 we want to find a time to take the motorcycle to reach a speed of 100
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The motorbike reaches 100 km/h in 3.5 seconds

Explanation:

The motion of the motorbike is a uniformly accelerated motion (= constant acceleration), therefore we can use the following suvat equation:

v=u+at

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v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the motorbike in this problem,

u = 0 (it starts from rest)

v = 100 km/h = 27.8 m/s is the final velocity

a=7.9 m/s^2 is the acceleration

Solving for t, we find the time it takes for the bike to reach that velocity:

t=\frac{v-u}{a}=\frac{27.8-0}{7.9}=3.5 s

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6 0
3 years ago
Find the weight of an object of mass 5 kg
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Answer:

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3 0
3 years ago
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Answer:

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Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.
Fudgin [204]

Answer:

g'_h=1.096\times 10^{-5}\ m.s^{-2}

Explanation:

We know that the gravity on the surface of the moon is,

  • g'=\frac{g}{6}
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<u>Gravity at a height h above the surface of the moon will be given as:</u>

g'_h=\frac{G.m}{(r+h)^2} ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

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  • h=384.4\times 10^6\ m is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}

g'_h=1.096\times 10^{-5}\ m.s^{-2} is the gravity on the moon the earth-surface.

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