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Damm [24]
3 years ago
9

In noisy factory environments, it's possible to use a loudspeaker to cancel persistent low-frequency machine noise at the positi

on of one worker. The details of practical systems are complex, but we can present a simple example that gives you the idea. Suppose a machine 5.5 m away from a worker emits a persistent 90 Hz hum. To cancel the sound at the worker's location with a speaker that exactly duplicates the machine's hum, how far from the worker should the speaker be placed? Assume a sound speed of 340 m/s.
Physics
1 answer:
Vaselesa [24]3 years ago
7 0

Answer:

7.39 m or 3.61 m

Explanation:

\lambda = Wavelength

f = Frequency = 90 Hz

v = Speed of sound = 340 m/s

Path difference of the two waves is given by

s_1-s_2=\frac{\lambda}{2}

Velocity of wave

v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{340}{90}\\\Rightarrow \lambda=3.78\ m

s_1=s_2\pm\frac{\lambda}{2}\\\Rightarrow s_1=5.5\pm \frac{3.78}{2}\\\Rightarrow s_1=7.39\ m, 3.61\ m

So, the location from the worker is 7.39 m or 3.61 m

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a shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. th
Savatey [412]

Answer:

15.7 m

Explanation:

The range (horizontal distance) of the projectile is determined only by its horizontal motion.

The horizontal motion is a motion with constant speed, which is equal to the initial horizontal velocity of the object:

v_x = v cos \theta

where

v = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle between the direction of v and the horizontal

Substituting,

v_x = (12.0 m/s)(cos 51.0^{\circ} )=7.55 m/s

We know that the projectile hits the ground in a time of

t = 2.08 s

so the horizontal distance covered is

d = v_x t = (7.55 m/s)(2.08 s)=15.7 m

8 0
2 years ago
If the box weighs 40 newtons and is lifted a distance of 2 meters, how much work is done on the box?
Zarrin [17]

Answer:

The work done on the box is 80 J.

Explanation:

Given that,

Weight of box = 40 N

Distance = 2 meter

We need to calculate the work done

Using formula of work done

W=F\times x

W=mg\times x

Where, x = distance

mg = weight

Put the value into the formula

W=40\times2

W= 80\ Nm

W=80\ J

Hence, The work done on the box is 80 J.

5 0
3 years ago
A 2.0-kg object moving at 5.0 m/s encounters a 30-Newton restive force
sveta [45]

The impulse experienced by the object is 3 N s.

<u>Explanation:</u>

Impulse is also termed as change in the momentum of the object. So, it is directly proportional to the force acting on the object and the time for which the force is acting on that object.

Thus, impulse experienced by an object is the product of force acting on the object for a given time period. So, it is the sudden influence of force on the given volume.

As the force is given as 30 N and the duration or the time is given as 0.1 seconds. Then, the impulse will be product of force with duration.

Impulse = Force × ΔTime = Force × Duration

Impulse = 30 × 0.1 = 3 N s.

Thus, the impulse experienced by the object is 3 N s.

6 0
3 years ago
What is happening when two waves with identical crests and troughs meet?
GenaCL600 [577]

Answer:

When two waves meet in such a way that their crests line up together, then it's called constructive interference. The resulting wave has a higher amplitude. In destructive interference, the crest of one wave meets the trough of another, and the result is a lower total amplitude.

7 0
2 years ago
What is the pressure exerted by an elephant who weighs 2500N and whose feet cover an area of 7.5m2?
qaws [65]

Answer:

0.00339905404

Explanation:

kg: 0.00339905404

pascals: 333.333333

pounds: 0.0483459126

8 0
3 years ago
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