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Julli [10]
3 years ago
11

How are musical instruments grouped?

Physics
2 answers:
vovikov84 [41]3 years ago
7 0

Musical instruments are grouped by how they produce the pitch.

Pitch is the highness or lowness of a tone.

rosijanka [135]3 years ago
6 0

A and B are equivalent.  That's one way instruments are often grouped. (the "sopranos", the "altos", the "bass")

C is another way instruments are often grouped; (the "woods", the "brass")

D is another way instruments are often grouped; (the "strings", the "percussions")

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Answer:

yes

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its on edg.

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Jennifer runs 5 miles east, then stops to take a break. after her break, she continues running 4 more miles east. what is her di
andreev551 [17]

Jennifer runs 5 miles east, then stops to take a break. after her break, she continues running 4 more miles east.

In this the total distance she covers is 9 miles and the total displacement is also the same as the distance that is 9 miles.

<h3>What is the difference between distance and displacement?</h3>

Displacement is the shortest distance between initial and final position, or we can say it is the straight-line distance between initial and final position.

Whereas distance is considered as the total path length covered from initial position till the final position. The Displacement of a body is always less than or equal to the distance.

Displacement can be zero in case the initial and final positions coincide, but distance can never be zero.

To know more about distance and displacement, visit:

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2 years ago
the atoms in a sample are close together but can slide past one another. as the atoms lose energy, they move slower. the atoms b
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The change of state that is occurring is from the liquid state to the solid state.
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3 years ago
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For which question could a testable hypothesis be developed?
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8 0
3 years ago
A basketball player is 4.22 m from
max2010maxim [7]

Answer: The height above the release point is 2.96 meters.

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A = (0, -9.8m/s^)

For the velocity we can integrate over time and get:

V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))

for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)

P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)

now, the time at wich the horizontal displacement is 4.22 m will be:

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t = (4.22/ 9.20*cos(69°)) = 1.28s

Now we evaluate the y-position in this time:

h =  -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m

The height above the release point is 2.96 meters.

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