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Ann [662]
3 years ago
12

Two loudspeakers (A and B) are 3.20m apart and emitting a sound with a frequency of 400Hz. An observer is 2.10m directly in fron

t of A. If the speed of sound in this room is 340m/s will the observer hear a loud sound or a quiet sound?
Physics
1 answer:
TiliK225 [7]3 years ago
5 0

Answer:

The observer hears a loud sound

Explanation:

In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.

You first calculate the distance between the observer and the loudspeakers.

The distances are given by:

d1: distance to loudspeaker A = 2.10m

d2: distance to loudspeaker B

d_2=\sqrt{(3.20m)^2+(2.10m)^2}=3.827m

Next, you calculate the wavelength of the sound waves by using the following formula:

\lambda=\frac{v_s}{f}

vs: speed of sound =  343 m/s

f: frequency of the waves = 400Hz

λ: wavelength

\lambda=\frac{343m/s}{400Hz}=0.8575m

Next, you calculate the path difference between the distance from the observer to the loudspeakers:

\Delta d=3.827m-2.10m=1.727m

You obtain a constructive interference (loud sound) if the quotient between the wavelength of the sound and the difference path is an integer:

\frac{\Delta d}{\lambda}=\frac{1.727m}{0.857}\approx2

Then, there will be a constructive interference, and the sound who the observer hears is loud.

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Newton's law of universal gravitation states that a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers
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A skier of mass 60 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 75 m up a 30° slop
postnew [5]

Answer:

Explanation:

Given

mass of skier=60 kg

distance traveled by skier=75 m

inclination(\theta )=30^{\circ}

speed (v)=2.4 m/s

as the skier is moving up with a constant velocity therefore net force is zero

F_{net}=0

Force applied by cable=mg\sin \theta

F=60\times 9.8\times \sin (30)=294 N

work done=F\cdot x

W=294\cdot 75=22.125 J

(b)Power=F\cdot v

P=294\cdot 2.4=705.6 W\approx 0.946 hp

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3 years ago
A car has a kinetic energy of 1.9 × 10^3 joules. If the velocity of the car is decreased by half, what is its kinetic energy?
VLD [36.1K]
The initial kinetic energy of the car is
E_1 =  \frac{1}{2}mv_1^2 =  1.9 \cdot 10^3 J

Then, the velocity of the car is decreased by half: v_2 =  \frac{v_1}{2}
so, the new kinetic energy is
E_2 =  \frac{1}{2}mv_2 ^2 =  \frac{1}{2} m ( \frac{v_1}{2} )^2= \frac{1}{2}m \frac{v_1^2}{4}= \frac{E_1}{4}
So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
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A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the no
padilas [110]

Explanation:

The weight of the car is equal to, W_c=m\times g...........(1)

Where

m is the mass of car

g is the acceleration due to gravity

The normal or vertical component of the force is, F_N=mg\ cos\theta

or

F_N=mg\ cos(13).............(2)

The horizontal component of the force is, F_H=mg\ sin\theta

Taking ratio of equation (1) and (2) as :

\dfrac{F_N}{W_c}=\dfrac{mg\ cos\theta}{mg}

\dfrac{F_N}{W_c}=cos(13)

\dfrac{F_N}{W_c}=0.97

or

\dfrac{F_N}{W_c}=\dfrac{97}{100}

Hence, this is the required solution.

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