an object is more dense if there is more weight in the same area, its because of the amount of particles and the weight of the particles. eg, 1cm cubed of gold would weigh alot more than 1cm cubed polystyrene, so gold is more dense

3 0

3 years ago

Two in-phase loudspeakers emit identical 1000 Hz sound waves along the x-axis. Part A What distance should one speaker be placed

Fiesta28 [93]

The distance a speaker should be placed behind other sound to have an amplitude 1.50 times is 4.523 m.

<h3>What is wavelength?</h3>

The wavelength is the distance between the adjacent crest or trough of the sinusoidal wave. The wavelength is the reciprocal of the frequency of the wave.

Wavelength λ = v/f

Two in-phase loudspeakers emit identical 1000 Hz sound waves along the x-axis.

λ = 343 m/s /1000 Hz

λ = 0.343 m

Distance, one should speaker be placed behind the other for the sound to have an amplitude 1.50 times that of each speaker alone.

The amplitude of the waveform due to waves,

A = 2a cos (ΔΦ/2)

ΔΦ = 2π x Δx/λ

So, A = 2a cos (π x Δx/λ)

Substitute the values, we get

1.5a = 2a cos (3.14 x Δx/ 0.343)

Δx = 4.523 m

Thus, the distance is 4.523 m.

Learn more about wavelength.

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7 0

2 years ago

42,43 and 44 please

Mariana [72]

42) The sailboat travels east with velocity $v_e=30 mph$ , while the current moves south with speed $v_s=30 mph$ . Since the two velocities are perpendicular to each other, he resultant velocity will be given by the Pytagorean theorem:
$v= \sqrt{v_e^2+v_s^2}= \sqrt{(30)^2+(30)^2}=42 mph$
and the direction is in between the two original directions, therefore south-east. So, the correct answer is
D) 42 mph southeast

43) Since the light moves by uniform motion, we can calculate the distance corresponding to one light year by using the basic relationship between velocity, space and time. In fact, we know the velocity:
$v=300,000,000 m/s=3 \cdot 10^8 m/s$
and the time is one year, corresponding to:
$t=32,000,000 s=3.2 \cdot 10^7 s$
therefore, the distance corresponding to one light year is:
$S=vt=(3\cdot 10^8 m/s)(3.2 \cdot 10^7 s)=9.6 \cdot 10^{15}m$
Therefore, the correct answer is D.

44) For the purpose of the problem, we can assume that the light travels instantaneously from the flash to us (because the distances involved are very small), so the time between the flash and the thunder corresponds to the time it took for the sound to travel to us.
The speed of sound is
$v=1100 ft/s=750 mph=335 m/s$
And since the time between the flash and the thunder is t=3 s, the distance is
$S=vt=(335 m/s)(3 s)=1005 m=0.62 miles$
Therefore, the correct answer is A) 3/5 mile.

6 0

3 years ago