Answer:
The frequency of the piano string is either 1053 HZ or 1059 HZ.
Explanation:
Here we know that frequency of beats is equal to the difference between the frequencies between two waves .
Given that frequency of tuning fork is 1056 HZ .
Let the frequency of the piano be ' f ' .
Given that number of beats = 3.
We know that | 1056 - f | = 3 ;
⇒ 1056- f = ±3,
Upon solving this we get
f = 1056-3 and 1056 + 3
⇒ f = 1053 or 1059 .
Answer: Gases have three characteristic properties: (1) they are easy to compress, (2) they expand to fill their containers, and (3) they occupy far more space than the liquids or solids from which they form. An internal combustion engine provides a good example of the ease with which gases can be compressed.
Explanation:
Answer: The volume is: " 1.430 cm³ " .
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Explanation:
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Volume, " V = ? ; (unknown, we need to solve for this).
Density, "D = 19.32 g / cm³ ;
mass , "m" = 27.63 g ;
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The formula for density is:
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D = m / V ; Divide each side of the equation by: "(1/m)" ;
to isolate " V" on one side of the equation; and to solve for "V" ;
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1/m)*D = (1/m) * (m / V) ;
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to get:
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D/m = ; ↔ 1 / V ; Take the reciprocal of EACH SIDE; to isolate "V" on each side of the equation:
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m / D = V/1 ↔ V = m / D ;
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V = m / D ;
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Now, plug in our given values for mass, "m" ; and Density, D"; to solve for "Volume, V " ;
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V = m / D = (27.63 g) ÷ (<span>19.32 g / cm</span>³) ;
= (27.63 g) * (1 cm³ / 19.32 g) ;
= (27.63 ÷ 19.32) cm³ ;
= 1.4301242236024845 cm³ ;
→ Round to "4 significant figures" ;
= 1.430 cm³ .
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The volume is: " 1.430 cm³ " .
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Answer:

Explanation:
In order for the car does not slip, the frictional force must be equal to the centripetal force due to the circular motion. According to the free body diagram:

The frictional force is given by:

The centripetal force is defined as:

Here v is the linear speed and r is the radius of the circular motion. Replacing this equations:

Answer:
gets higher
Explanation:
There are videos that show the range of human hearing. If you would play the video, you would notice that if the frequency increases, the pitch would also increase.