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3 years ago

Why does the midi system have the widest range of frequencies

2 answers:

7 0

<span>Question:
Why does the midi system have the widest range of frequencies

Answer:
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Each of these instruments have a very wide range of notes because these are all treble instruments. Now, if you mean altogether, with a trumpet you can hit a very wide range of notes due to it having such a high octave, same with a violin, piano, clarinet, and organ. Kettle drums just have a very manipulable sound.
But remember it is a MIDI System, and you can adjust the frequency according to your desire; So, honestly, there is no straight answer for this, just how you desire the sound.

I could take an A4 note on a trumpet and stretch the sound waves out to sound like an A1 note just by bending it with a MIDI System. So, the possibilities are endless.

Hope this helps :D

P.S been playing the piano for 7 years so I use midi alot.

Mrac [35]3 years ago

5 0

The Midi System has the widest range of frequencies because it doesn't care about the range of frequency of the actual instrument it is playing. it is a event oriented system.

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I will mark brainiest! Please help ASAP

Lera25 [3.4K]

Answer:

Every substance is made up of tiny units called atoms. Each atom has electrons, particles that carry electric charges. Spinning like tops, the electrons circle the nucleus, or core, of an atom. Their movement generates an electric current and causes each electron to act like a microscopic magnet

Explanation:

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3 years ago

For the magnetic field at some random angle to the plane of the small coil, draw a picture showing only the small coil, a vector

bearhunter [10]

Answer:

Solution attached below

Explanation:

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4 years ago

Consider a thin rod of mass 3.2 kg, length 1.2 m and uniform density. The rod is pivoted at one end on a frictionless horizontal

notsponge [240]

Answer:

the angular acceleration is 9.7 rad/ $s^{2}$

Explanation:

given information:

mass of thin rod, m = 3.2 kg

the length of the rod, L = 1.2

angle, θ = 38

to find the acceleration of the rod, we can use the torque's formula as below,

τ = Iα

where

τ = torque

I = inertia

σ = acceleration

moment inertia of this rod, I

I = $\frac{1}{3} mL^{2}$

τ = F d, d = $\frac{L}{2}$ cosθ

τ = m g $\frac{L}{2}$ cosθ

now we can substitute the both equation,

τ = Iα

α = τ/I

= (m g $\frac{L}{2}$ cosθ)/( $\frac{1}{3} mL^{2}$ )

= 3gcosθ/2L

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3 0

3 years ago

Kinetic energy and mass are ________ proportional.

omeli [17]

Answer:

The Kinetic energy and mass are _directly_ proportional.

Explanation:

We know that Kinetic Energy is basically termed as the capacity of a body to do work.

Kinetic energy is often used to associate with moving objects, therefore, K.E is normally termed as the energy of motion.

The formula of K.E of an object of mass and velocity is defined

K.E = 1/2mv²

From the formula, it is clear that K.E is directly proportional to its mass and also directly proportional to the square of its velocity.

For example,

If A toy plane with a mass of 10 kg is flying at 20 m/s. Its K.E will be:

m = 10 kg

v = 20 m/s

K.E = 1/2mv²

= 1/2(10)(20)²

= 1/2(10)(400)

= 5(400)

= 2000 J

Now, let suppose, if we double the mass of a toy plane i.e.

m = 20 kg

K.E = 1/2mv²

= 1/2(20)(20)²

= 1/2(20)(400)

= 10(400)

= 400 J

Therefore, the K.E is doubled when doubled the mass.

Therefore, the Kinetic energy and mass are _directly_ proportional.

9 0

3 years ago

Read 2 more answers

Two long, parallel transmission lines, 40.0cm apart, carry 25.0-A and 73.0-A currents.A). Find all locations where the net magne

In-s [12.5K]

Answer:

a) If the currents are in the same direction, the magnetic field is zero at x = 0.298 m = 29.8 cm

That is, in between the wires, 29.8 cm from the 73.0 A wire and 10.2 cm from the 25.0 A wire.

b) If the currents are in opposite directions, the magnetic field is zero at x = 0.608 m = 60.8 cm

That is, along the positive x-axis, 60.8 cm from the 73.0 A wire and 20.8 cm from the 25.0 A wire.

Explanation:

The origin is at the 73.0 A wire and the 25.0 A wire is at x = 0.40 m

The magnetic field in a current carrying wire at a distance r from the wire is given by

B = (μ₀I/2πr)

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

a) If the currents are in the same direction, at what positions is the magnetic field equal to 0.

According to laws describing the direction.of magnetic fields, this position will be at some point between the two wires.

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire points into the plane of the book, moving in the negative x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (0.4 - x)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(0.4-x)]

(73/x) = [25/(0.4-x)]

73(0.4-x) = 25x

29.2 - 73x = 25x

73x + 25x = 29.2

98x = 29.2

x = (29.2/98) = 0.298 m

b) If the currents are in the opposite directions, at what positions is the magnetic field equal to 0?

According to laws describing the direction.of magnetic fields, this position will be at some point beyond the second wire (since we're initially concerned about the positive x-direction).

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire (whose direction is now in the opposite direction to the current in the first wire) is also along the positive x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (x - 0.4)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(x-0.4)]

(73/x) = [25/(x-0.4)]

73(x-0.4) = 25x

73x - 29.2 = 25x

73x - 25x = 29.2

48x = 29.2

x = (29.2/48) = 0.608 m

Hope this Helps!!!

5 0

4 years ago