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LiRa [457]
3 years ago
5

If you wanted the pitch of a horn to drop relative to an observer, which way would you move the horn, relative to where that obs

erver is located
Physics
1 answer:
Vladimir [108]3 years ago
3 0
We assume that horn releases sound of constant frequency. In order for observer to observe different frequency either horn or observer or both must move.

This happens due to Doppler effect. It states that when position of source of sound and observer relative to each other changes, the observed frequency also changes. If the source emits sound of constant frequency than observed frequency will be either higher or lower than original.

When distance between source and observer increases the observed frequency will be lower. This is because same number of sound waves must cover greater distance so they have greater wavelength.
When distance between source and observer decreases the observed frequency will be higher. This is because same number of sound waves must cover smaller distance so they have smaller wavelength. 

Wavelength and frequency are inversely proportional meaning when one increases the other drecreases.

From this explanation we can find answer for our question. <span>If we wanted the pitch of a horn to drop relative to an observer we need to move horn away from an observer.</span>
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Please help me with this question. BRAINLIEST if answered.
IceJOKER [234]

... The top branch of the 3-branched parallel block ... the 9 and 6 in series ...
is equivalent to a single resistor of 15 ohms.

... The 3-branched parallel block boils down to (30, 10, and 15) in parallel.
That's (1/30 + 1/10 + 1/15)⁻¹  =   5 ohms.

... The 5-ohm-equivalent block and the 20-ohm resistor form a
voltage divider across the battery.
The voltage across the 5-ohm-equivalent block is (5/25 x 30v) = 6v .

... The top branch of the block is equivalent to a (9 + 6) = 15-ohmer.
With 6v across its ends, the current through that branch is (6/15) = 0.4A .

... With 0.4A flowing through it, the 9-ohm resistor is dissipating

            I²R = (0.4A)² (9 ohms)  =  (0.16 A²) (9 ohms)  =  1.44 W  (choice-3)   
8 0
3 years ago
Waves begin to "feel bottom" when the depth of water is
LekaFEV [45]
I think the correct answer would be one half the wavelength. Waves would "feel bottom" when the water is at the depth of 0.5 of the wavelength. "Feel bottom" is a term used to describe that the depth of water affects the wave properties. Hope this answers the question.
8 0
3 years ago
Read 2 more answers
Can I improve the design of my simple machine? How?
blsea [12.9K]

Answer:

12345

Explanation:

yan na po answer ko hehehe

5 0
3 years ago
An air conditioner operating at steady state maintains a dwelling at 70oF on a day when the outside temperature is 90oF. If the
torisob [31]
30000 btuh /3413 btuh/kw. = 8.8 kw

8.8 kw/.746 kw/hp = 11.8 hp if COP is 1

11.8/3 hp (COP coefficient of performance) = 3.99 COP

>>>So yes a 3.0 hp compressor with a nominal COP of 4 will handle the 30,000 btuh load.

3.2 to 4.5 is expected COP range for an air cooled heat pump or a/c unit.
8 0
3 years ago
A body in simple harmonic motion has a displacement x that varies in time t according to the equation x = 5cos(π t + π/3) , wher
zhuklara [117]

Answer:

1/2 Hz

Explanation:

A simple harmonic motion has an equation in the form of

x(t) = Acos(\omega t - \phi)

where A is the amplitude, \omega = 2\pi f is the angular frequency and \phi is the initial phase.

Since our body has an equation of  x = 5cos(π t + π/3) we can equate \omega = \pi and solve for frequency f

2\pi f = \pi

f = 1/2 Hz

7 0
3 years ago
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