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Butoxors [25]
3 years ago
11

1. When playing the piano, each key has a specific number of times it vibrates. The number of string vibrations correspond to ch

anging what quality of sound? loudness
pitch
amplitude
range
2. What is the lowest level of sound that can be detected by human hearing?
0 Hz
10 Hz
20 Hz
100 Hz
3.

The sound waves used for sonar are _____.
infrasound
ultrasound
audible
4. At close range, a ship's foghorn or the air horn on a truck will probably have a ______(High, Low, Medium) decibel reading.
5. A loud explosion will produce _____.
a high-amplitude sound wave
a low-amplitude sound wave
a sound wave with a single frequency
Physics
2 answers:
kakasveta [241]3 years ago
3 0

Answer:

2: 20HZ 3:audiable

Explanation:

i guessed and it told me i was right

PilotLPTM [1.2K]3 years ago
3 0

Answer: Hello mate!

1) the number of times that the string vibrates ober the unit of time, is the frequency of the note.

This canges the pitch of the note, where a high frequency is asociated with a high pitch (the first string in the guitar) and a low frequency is asociated with a low pitch (the thickest string on a guitar)

2) Is commonly stated that the human hearing can detect sounds of 20Hz, but in some situations can be detected sounds with lower frequency, where the samallest detected is 12Hz

3) The sound waves used for a sonar can vary from infrasound to ultrasound, depending of the range that the sonar covers, there is no right answer here.

4) A high decibel reading, this is because the source "shots" the sound un a prefered direction, then the intensity of the sound is very high right next to the source.

5) A loud explosion will produce a high-amplitude sound wave, this is because the amplitud is directly relationated with the intensity (or loudnes) of the sound wave.

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The H line in Calcium is normally at 396.9nm. however, in a star’s spectrum it is measured at 398,1 nm. How fast is the star mov
konstantin123 [22]

H line of Calcium spectrum is normally given as 396.9 nm

Now from a distant star we measured it as 398.1 nm

So here this change in the wavelength of distant star is due to Doppler's effect of light as per which when source of light moves towards the observer then the frequency of light received will appear different from its actual frequency

So here we can say as per Doppler's effect of light

\frac{\Delta \nu}{\nu_0} = \frac{v}{c}

\frac{\nu' - \nu}{\nu_0} = \frac{v}{c}

\frac{\frac{1}{\lambda} - \frac{1}{\lambda'}}{\frac{1}{\lambda}} = \frac{v}{c}

\frac{\lambda' - \lambda}{\lambda'} = \frac{v}{c}

given that

\lambda' = 398.1 nm

\lambda = 396.9 nm

c = 3 * 10^8 m/s

\frac{398.1 - 396.9}{398.1} = \frac{v}{3*10^8}

v = 3*10^8 * \frac{1.2}{398.1}

v = 9.04 * 10^5 m/s

so the start is moving away with speed 9.04 * 10^5 m/s because when wavelength is more than the real wavelength then its frequency is less which mean it is moving away from the Earth

5 0
3 years ago
A golf ball with an initial angle of 34° lands exactly 240 m down the range on a level course.
g100num [7]

Answer:50.39 m/s,

40.46 m

Explanation:

Given

launch angle=34^{\circ}

Range=240 m

We know that Range =\frac{u^2sin2\theta }{g}

240=\frac{u^2sin(68)}{9.81}

u=50.39 m/s

(b)maximum height of projectile is given by

H=\frac{u^2sin^2\theta }{2g}

H=\frac{50.39^2(sin34)^2}{2\times 9.81}

H=40.46 m

3 0
4 years ago
Read 2 more answers
If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind th
Sedbober [7]

Answer:

c = 4,444.44

Explanation:

You have the following expression for the acceleration of the projectile:

a=cs   (1)

s: distance to the ground of the projectile

To find the value of the constant c you use the following formula:

v^2=v_o^2+2a \Delta s   (2)

vo: initial  velocity = 0 m/s

v: final speed = 200 m/s

Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m

You replace the expression (1) into the expression (2):

v^2=2(cs)\Delta s

You do the constant c in the last equation, then you replace the values of v, s and Δs:

c=\frac{v^2}{2s\Delta s}=\frac{(200m/s)^2}{2(3m/s^2)(1.5m)}=4444.44

6 0
3 years ago
If the change in kinetic energy of a tennis ball hit by the racket
Marrrta [24]

Answer: .36 m

Explanation:

5 0
3 years ago
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jasenka [17]

Answer:

She pulled the scarf from her neck and wiped her face.

4 0
3 years ago
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