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NARA [144]
2 years ago
6

Which observation indicates a chemical property?

Physics
2 answers:
koban [17]2 years ago
6 0

Answer:

Turns black.

Explanation:

nikklg [1K]2 years ago
5 0
Slowly desovles i think...
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A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she
lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, L_{new} = 2/3·L

The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0
3 years ago
How much would it cost to cover the entire land area of the U.S. in dollar bills?
agasfer [191]

Answer:

  $900 trillion

Explanation:

If Alaska is 20% of the contiguous US, then the approximate area of interest is ...

  1200 miles × 3000 miles = 3.6×10^6 square miles.

The size of a dollar bill is about ...

  (6.5 cm)·(15.5 cm) = 100.75 cm^2

One mile is 160,934.4 cm, so 1 square mile is about ...

  1 mi^2 = (160,934.4 cm)^2 ≈ 2.59·10^10 cm^2

The number of dollars of interest is then ...

  (3.6 · 10^6 mi^2)(2.59 · 10^10 cm^2)/(100.75 cm^2) ≈ 9.3·10^14

  ≈ 930 × 10^12 . . . dollars

It would cost about 900 trillion dollars to cover the land area of the US in $1 bills.

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3 years ago
A book is resting on the table. Forces are exerted _____.
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