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2 years ago

How can we alter the pitch of guitar

Physics

2 answers:

Vikentia [17]2 years ago

8 0

Explanation:

Guitar strings are tuned (tightened and loosened) using their tuning keys. Applying too much tension to a string tightly can raise it to the pitch of the next note, while loosening it can easily lower it the same amount. Increasing the tension raises the pitch. The length of a string is also important.

zhenek [66]2 years ago

4 0

Answer:

Tuning the keys.

Explanation:

"Guitar strings are tuned (tightened and loosened) using their tuning keys. Applying too much tension to a string tightly can raise it to the pitch of the next note, while loosening it can easily lower it the same amount. Increasing the tension raises the pitch. The length of a string is also important."

I hope this answers your question! Have a good night!

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Which value is equivalent to 178 centimeters?

IRINA_888 [86]

178 centimeters=1.78 meters
which means that the answer is C.

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2 years ago

U. A hockey player takes a slap shot at a puck at rest on

faltersainse [42]

Answer:

a) 7200 ft/s²

b) 140 ft

c) 3.7 s

Explanation:

(a) Average acceleration is the change in velocity over change in time.

a_avg = Δv / Δt

We need to find what velocity the puck reached after it was hit by the hockey player.

We know it reached 40 ft/s after traveling 90 feet over rough ice at an acceleration of -20 ft/s². Therefore:

v² = v₀² + 2a(x − x₀)

(40 ft/s)² = v₀² + 2(-20 ft/s²)(100 ft − 10 ft)

v₀² = 5200 ft²/s²

v₀ = 20√13 ft/s

So the average acceleration impacted to the puck as it is struck is:

a_avg = (20√13 ft/s − 0 ft/s) / (0.01 s)

a_avg = 2000√13 ft/s²

a_avg ≈ 7200 ft/s²

(b) The distance the puck travels before stopping is:

v² = v₀² + 2a(x − x₀)

(0 ft/s)² = (5200 ft²/s²) + 2(-20 ft/s²)(x − 10 ft)

x = 140 ft

t = (10 ft) / (20√13 ft/s)

t = (√13)/26 s

The time the puck travels over the rough ice is:

v = at + v₀

(0 ft/s) = (-20 ft/s²) t + (20√13 ft/s)

t = √13 s

So the total time is:

t = (√13)/26 s + √13 s

t = (27√13)/26 s

t ≈ 3.7 s

8 0

2 years ago

An investigation has been completed similar to the one on latent heat of fusion, where steam is bubbled through a container of w

allochka39001 [22]

Answer:

$Q_a=330 cal$

$Q_w=6000cal$

Explanation:

From the question we are told that

Mass of the aluminum container 50 g

Mass of the container and water 250 g

Mass of the water 200 g

Initial temperature of the container and water 20°C

Temperature of the steam 100°C

Final temperature of the container, water, and condensed steam 50°C

Mass of the container, water, and condensed steam 261 g

Mass of the steam 11 g Specific heat of aluminum 0.22 cal/g°C

a) Heat energy on container

Generally the formula for mathematically solving heat gain

$Q_c=M_c *C_c*( \triangle T)$

Therefore imputing variables we have

$Q_a=50g *0.22*50-20$

$Q_a=330 cal$

b) Heat energy on water

Generally the formula for mathematically solving heat gain

$Q_w=M_w *C_w*( \triangle T)$

Therefore imputing variables we have

$Q_w=200 *1* 50-204$

$Q_w=6000cal$

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3 years ago

What is the SI unit for momentum?

serg [7]

It’s a vector quantity, which means it possesses both magnitude and direction. So the SI unit would be B)kg•m/s

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2 years ago

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What is the magnitude of the velocity when the elastic potential energy is equal to the kinetic energy? (Assume that U=0 at equi

zhannawk [14.2K]

Answer:

Explanation:

General Equation of SHM is given by

$x=A\cos \omega t$

$v=-A\omega \sin \omega t$

where x=position of particle

A=maximum Amplitude

$\omega =$ angular frequency

t=time

At any time Total Energy is the sum of kinetic Energy and Elastic potential Energy i.e. $\frac{1}{2}kA^2$

where k=spring constant

Potential Energy is given by $U=\frac{1}{2}kx^2$

also it is given that Potential Energy(U) is equal to Kinetic Energy(K)

Total Energy $=K+U$

Total $=2U=2\times \frac{1}{2}kx^2$

$\frac{1}{2}kA^2=2\times \frac{1}{2}kx^2$

$x=\pm \frac{A}{\sqrt{2}}$

at $x=\frac{A}{\sqrt{2}}$

velocity is $v=\frac{A\omega}{\sqrt{2}}$

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3 years ago