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Zigmanuir [339]
3 years ago
6

What is the central peak in the central Arzachel

Physics
1 answer:
Tema [17]3 years ago
5 0

Answer:

The rugged central peak of Arzachel is prominent, rising 1.5 kilometers above the floor, and is somewhat offset to the west with a bowed curve from south to north-northeast. The floor is relatively flat, except for some irregularities in the southwestern quadrant of the crater.

Explanation:

The rugged central peak of Arzachel is prominent, rising 1.5 kilometers above the floor, and is somewhat offset to the west with a bowed curve from south to north-northeast. The floor is relatively flat, except for some irregularities in the southwestern quadrant of the crater.

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A flutist assembles her flute in a room where the speed of sound is 342m/s . When she plays the note A, it is in perfect tune wi
USPshnik [31]

Answer:

a.3Hz

b.0.0034m

Explanation:

First, we know the flute is an open pipe, because open pipe as both end open and a close organ pipe as only one end close.

The formula relating the length and he frequency is giving as

f=\frac{nv}{2l}\\.

a.we first determine the length of the flute at the fundamental frequency i.e when <em>n</em>=1 and when the speed is in the 342m/s

Hence from

f=\frac{nv}{2l}\\\\l=\frac{342}{2*440}\\ l=0.389m\\.

since the value of the length will remain constant, we now use the value to determine the frequency when the air becomes hotter and the speed becomes 345m/s.

f=\frac{nv}{2l} \\f=\frac{345}{2*0.389}\\f=443.4Hz

Hence the require beat is

B=/f_{1}-f_{2}/\\B=/440-443/\\B=3Hz.

b. since the length is dependent also on the speed and frequency, we determine the new length when she plays with a fundamental frequency when the speed of sound is 345m/s

using the formula

L_{new}=\frac{v}{2f}\\\\L_{new}=\frac{345}{2*440}\\\\L_{new}=0.39204

Now to determine the extension,

L_{extend}=L_{new}-L_{old}\\L_{extend}=0.39204- 0.38864\\L_{extend}=0.0034m\\

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