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Ganezh [65]
3 years ago
8

Two loudspeakers are placed side by side and driven by the same source at 500 Hz. A listener is positioned in front of the two s

peakers and on the line separating them, thus creating a constructive interference at the listener's ear. If one of the speakers is gradually pushed toward the listener, how far must it be moved to repeat the condition of constructive interference at the listener's ear
Physics
1 answer:
Oliga [24]3 years ago
4 0

Answer:

0.68 m

Explanation:

We know that the speed of sound in air is a product of frequency and wavelength. Taking speed of sound in air as 340 m/s

V=frequency*wavelength

Then wavelength is given by 350/500=0.68 m

Therefore, to repeat constructive interference at the listener's ear, a distance of 0.68 m is needed

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Suppose there was a star with a parallax angle of 1 arcsecond. How far away would it be?
jok3333 [9.3K]

Answer: 3.26 light years

Explanation:

Each star has a parallax of one arcsecond at a distance of one parsec, which is equivalent to 3.26 light years.

so the parallax of 1 arcsecond will be at a distance of <em>1/1 × 3.26 light years</em>

3 0
3 years ago
Consider a system two point charges. One has charge +q at (x, y,z) -(a,0,0) and another of charge-q at (x, y, z) = (-a, 0,0). 5.
olga2289 [7]

Answer:

electricfield at (0,0,0) is Et = 2 k q / a²

Explanation:

For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity

For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation

     E = k q / r²

Point (0,0,0)

We calculate for the charge -q which is at a distance R = a

   E1 = k (-q) / a²

   E1 = - kq / a²

As the test charge is positive in the field it goes to the left, attractive force

We calculate for the charge that is also at R = a

    E2 = k q / a²

This field goes to the left, repulsive force

We find the total electric field

    Et = E1 + E2

    Et = kq / a² + kq / a²

    Et = 2 k q / a²

Point (0,0, R)

We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a

     E1 = k q / (R + a)²

     E2 = kq / (R-a)²

     Et = kq [1 / (R + a)² + 1 / (R-a)²]

     Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}

     Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}

If we use the condition that  R> a we can despise in the patents "a"

     (R² + a²) = R² (1+ a² / R²) ≈ R²

     (R + a)² = R² (1 + a / R)² ≈ R²

     (R- a)²  = R² (1-a / R)² ≈ R²

Substituting in the total electric field

     Et = kq {2 R²) / [R²R²]}

     Et =kq 2 / R²

7 0
3 years ago
A brick of mass 1.15 kg is attached to a thin (massless) cable and whirled around in a circle in a vertical plane. The circle ha
stira [4]

Explanation:

It is given that,

Mass of the brick, m = 1.15 kg

Radius of the circle, r = 1.44 m

The cable will break if the tension exceeds 43.0 N

Let v is the maximum sped can have at the bottom of the circle before the cable will break. At the bottom of the circle, the net force is equal to the centripetal force along with the weight of the brick. So,

T=\dfrac{mv^2}{r}+mg

\dfrac{T}{m}-g=\dfrac{v^2}{r}

v=\sqrt{(\dfrac{T}{m}-g)r}

v=\sqrt{(\dfrac{43}{1.15}-9.8)\times 1.44}

v = 6.30 m/s

So, the maximum speed of the brick at the bottom of the circle before the cable will break is 6.3 m/s. Hence, this is the required solution.

8 0
3 years ago
A car is pushed with a force of 450 N for 19.4 seconds. What impulse was applied to the car?​
harina [27]

Answer:

impulse = 8820 kg·\frac{m}{s} or 8820 N·s

Explanation:

Impulse J is equal to the average force F_{av} multiplied by the elapsed time Δt or in equation form, J = F_{av}Δt

As long as your force of 450 N is constant then that value is your average force F_{av} and your elapsed time is 19.4 seconds.

Multiply these values.

You will get an impulse of 8820 kg·\frac{m}{s} or 8820 N·s.

6 0
2 years ago
To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
3 years ago
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