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zhannawk [14.2K]
1 year ago
13

Certain noteheads can only be placed on staff lines.

SAT
1 answer:
professor190 [17]1 year ago
8 0

The statement Certain noteheads can only be placed on staff lines is: "False.

<h3>What is noteheads?</h3>

Noteheads can be placed on staff lines or in the spaces between them. The position of the notehead determines the pitch of the note. A notehead on a line or space corresponds to a specific letter name (A, B, C, D, E, F, or G) and octave (the range of pitches from one C to the next C).

For example, in the treble clef, the notehead on the bottom line of the staff is an E, and the notehead in the space above it is an F. The notehead on the top line of the staff is an F, and the notehead in the space above it is a G.

Sometimes, noteheads need to be placed above or below the staff to indicate higher or lower pitches. In this case, ledger lines are used to extend the staff. A ledger line is a short horizontal line that acts as a staff line for a single note. A notehead can be placed on a ledger line or in the space between ledger lines.

For example, in the treble clef, the notehead on the first ledger line above the staff is an A, and the notehead in the space above it is a B. The notehead on the second ledger line above the staff is a C, and the notehead in the space above it is a D.

Here is an example of noteheads on staff lines, in spaces, and on ledger lines in the treble clef:

D

 ---

C |   |

 ---

B |   |

 ---

A |   |

 ---

G |   |

 ---

F |   |

 ---

E |   |

 ---

 E F G A B C D

"

Therefore the statement is false.

Learn more about Noteheads  here:brainly.com/question/29871741

#SPJ4

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A typical neutron star has a mass of about 1. 5msun and a radius of 10 kilometers
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How thick a layer would Earth form as it wraps around the neutron star’s surface is: 6.67 10⁻³ m.

<h3>Density of the Neutron star</h3>

Density

ρ = m / V

Where:

ρ= density

m = mass of the planet 5.98 10²⁴ km

V =volume of the spherical layer

Volume of a sphere

Volume = 4/3 π r³

Mass = 1.5 = 1.5 1,991 10³⁰

Mass= 2.99 10³⁰ kg

Density:

ρ = 2.99 10³⁰ / [4/3 π (10 10³)³]

ρ is = 7.13 10 17 kg / m³

V = 5.98 10²⁴ / 7.13 10¹⁷

V = 8,387 10⁶ m³

Thickness of the layer

V = 4π r² e

e = V / 4π r

e = 8,387 10⁶ / [4π (10 10³)²]

e = 6.67 10⁻³ m

Inconclusion how thick a layer would Earth form as it wraps around the neutron star’s surface is: 6.67 10⁻³ m.

Learn more about Density of the Neutron star here:brainly.com/question/15700804

3 0
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Ray purchases a paperweight and wants to cover it in paper. He creates the picture below to show the net of the figure. Ray only
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The square inches of paper Ray needs are equal to the area of the sides he wants to cover.

<h3>What is the area?</h3>

The area of an object refers to the surface or space occupied by the object.

<h3>Why is the area important?</h3>

In this situation calculating the area is essential because by doing this, Ray can know how much paper he needs.

<h3>How to calculate the area?</h3>

Different formulas can be used to calculate the area depending on the shape of the object.

  • Rectangle: Side x side
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This means to know the exact answer, Ray needs to calculate the area of the sides he wants to cover and this is equal to the paper he needs to buy.

Note: This question is incomplete because the picture was not attached. Due to this, I answered it based on general knowledge.

Learn more about area in: brainly.com/question/16151549

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Explanation:

     If you had the options:

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<em>Learn more about your problem here: brainly.com/question/678565</em>

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