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AlladinOne [14]
3 years ago
6

A standard frequency for tuning musical instruments is 440 Hz for the pitch of A, denoted A4. What is the length, in meters, of

the steel rod that produces the pitch A4 as its fundamental longitudinal resonance?
Physics
1 answer:
Sophie [7]3 years ago
4 0

Answer:

Explanation:

frequency of tuning fork = 440 Hz

wave length λ = velocity of sound  / frequency

= 343 / 440 m

= .7795 m

= 77.95 cm

λ = 77.95 cm

Let length of rod having fundamental wavelength of 77.95 cm be L

L = λ  /2

L = 77.95 / 2

= 38.975 cm .

length of rod = 38.975 cm .

= 0. 39 m

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Juliette [100K]

0.017s

Explanation:

Given parameter

Distance = 5000km = 5 x 10⁶m

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We use the normal speed equation to solve this problem as we take the speed of the wave to be 3 x 10⁸m/s.

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8 0
2 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin
lyudmila [28]
Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
W= \int\limits^{0.540m}_{0} {F} \, dx =  \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =
=16000x+10000  \frac{x^2}{2} - 26000  \frac{x^3}{3}
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
W=16000(0.540m)+10000  \frac{(0.540m)^2}{2} - 26000  \frac{(0.540m)^3}{3}  =
=8733 J=8.73 kJ

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
W=16000(0.95m)+10000  \frac{(0.95m)^2}{2} - 26000  \frac{(0.95m)^3}{3}  =
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5 0
2 years ago
A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w
Bumek [7]

Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

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mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

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B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

0.475v² = 0.0310644

v² = 0.0310644/0.475

v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

8 0
3 years ago
An electron is a particle with a ____.
Dahasolnce [82]

A beta particle. Hoped I help. Sorry if it wrong.

4 0
3 years ago
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Brut [27]
The magnitude of electric field is produced by the electrons at a certain distance.

E = kQ/r²

where: 
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Q = charge
r = distance
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Q = </span><span>1.602 × 10^–19 C
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8 0
3 years ago
Read 2 more answers
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