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3 years ago

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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.)

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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A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wir

Fantom [35]

Answer:

linear charge density = -9.495 × $10^{-34}$ C/m

Explanation:

given data

revolutions per second = 1.80 × $10^{6}$

radius = 1.20 cm

solution

we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force

so

- q × E = m × w² × r ..................1

- 9 × $10^{9}$ × $\frac{2*linear\ charge\ density}r}$ q = m × w² × r ............2

and w = $\frac{2*\pi}{T}$

w = $\frac{d\theta }{dt}$

w = 1.80 × $10^{6}$ × $\frac{2*\pi}{1}$

w = 11304000 rad/s

so here from equation 2

- 9 × $10^{9}$ × $\frac{2*linear\ charge\ density}{0.012}$ 1.80 × $10^{6}$ = 1.672 × $10^{-27}$ × 11304000² × 0.0120

linear charge density = -9.495 × $10^{-34}$ C/m

8 0

3 years ago

An electromagnet on the ceiling of an airplane holds a steel ball. When a button is pushed, the magnet releases the ball. The ex

blagie [28]

Answer:

The ball will fall on the X .

Explanation:

At height, when the aeroplane is in great speed , everything attached with it acquires the same speed . So ball will also have the same speed as the aeroplane have. When ball starts falling off , it gets detached from plane but , at the same time it continues to travel with its earlier speed , because of inertia of motion. So it remains stationary with respect to plane in horizontal direction . It has velocity with respect to plane only in vertical direction. Hence it will fall on the X. It is due to first law of motion.

7 0

3 years ago

An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where ther

jeka94

Answer:

T = 188.5 s, correct is C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

initial instant. Before the crash

L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

L_f = I₀ w + m r v

L₀ = L_f

m r v₀ = I₀ w + m r v

angular and linear velocity are related

v = r w

w = v / r

m r v₀ = I₀ v / r + m r v

m r v₀ = (I₀ / r + mr) v

v = $\frac{m}{\frac{I_o}{r} +mr} \ r v_o$

let's calculate

v = $\frac{0.020}{\frac{1.4}{0.6 } + 0.020 \ 0.6 } \ 0.6 \ 4$

v = $\frac{0.020}{2.345} \ 2.4$

v = 0.02 m / s

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

v = x / T

T = x / v

the distance of a circle with radius r = 0.6 m

x = 2π r

we substitute

T = 2π r / v

let's calculate

T = 2π 0.6/0.02

T = 188.5 s

reduce

t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is C

6 0

3 years ago

2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or

krek1111 [17]

a) The work done by the tree is $-1.44\cdot 10^5 J$

b) The amount of force applied is 2880 N

Explanation:

a)

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

$W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

$W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J$

The work is negative because it is done in the direction opposite to the direction of motion of the car.

b)

The work done by the tree on the car can also be rewritten as

$W=Fd$

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

$W=-1.44\cdot 10^5 J$ is the work done

$d=50.0 cm = 0.50 m$ is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

$F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N$

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0

3 years ago

A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle

Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of <em>θ</em> = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height <em>y</em> at time <em>t</em> would be

<em>y</em> = (40 m/s) <em>t</em> - 1/2 <em>g t</em> ²

Set <em>y</em> = 160 m, and you'll find that there is no (real) solution for<em> t</em>, so the ball never attains the given maximum height.

From another perspective: recall that

<em>v </em>² - <em>v</em>₀² = 2<em>a </em>∆<em>y</em>

where

• <em>v</em>₀ = initial velocity

• <em>v</em> = final velocity

• <em>a</em> = acceleration

• ∆<em>y</em> = displacement

At its maximum height, the ball has zero vertical velocity, and ∆<em>y</em> = maximum height = 160 m. The ball is in free fall once it's launched, so <em>a</em> = -<em>g</em>.

So we have

0² - (40 m/s)² = -2<em>g </em>(160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0

3 years ago