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RoseWind [281]
2 years ago
9

What's the largest number of beats per second that will be heard from which pair of tuning forks?

Physics
1 answer:
skelet666 [1.2K]2 years ago
7 0

The largest number of beats per second that will be heard from 763 and 774 Hz  pair of tuning fork .

The beat frequency is equal to the complete value of the alteration in the frequency of the two waves. The count of beats per second is equivalent to the difference in frequencies of two waves is called beat frequency.

A tuning fork is a two-pronged metal fork that can be used as an acoustic resonator. Traditionally, this tool has been used to tune musical instruments. Tuning forks work by releasing a perfect wave pattern to match a musician's instrument.

The largest number of beats per second that will be heard from 763 and 774 Hz  pair of tuning fork .

To learn more about beat frequency here

brainly.com/question/21923211

#SPJ4

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Determine the work done by the constant force. The locomotive of a freight train pulls its cars with a constant force of 15 tons
Digiron [165]

Answer:

5.92×10⁷ J

Explanation:

We'll begin by converting 15 tons to Newton. This can be obtained as follow:

1 ton = 9806.65 N

Therefore,

15 ton = 15 ton × 9806.65 N / 1 ton

15 ton = 147099.75 N

Next, we shall convert one-quarter (¼) or 0.25 mile to metre. This can be obtained as follow:

100 mi = 160934 m

0.25 mi = 0.25 mi × 160934 / 100 mi

0.25 mi = 402.335 m

Finally, we shall determine the Workdone. This can be obtained as follow:

Force (F) = 147099.75 N

Distance (d) = 402.335 m

Workdone (Wd) =?

Wd = F × d

Wd = 147099.75 × 402.335

Wd = 5.92×10⁷ J

Thus, the Workdone is 5.92×10⁷ J

5 0
2 years ago
. A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is orient
Dmitriy789 [7]

Answer:

The magnetic flux through surface is 2.22 \times 10^{-3} Wb

Explanation:

Given :

Magnitude of magnetic field B = 0.078 T

Radius of circle r = 0.10 m

Angle between field and surface normal \theta = 25°

From the formula of flux,

\phi = B.A

\phi = BA\cos \theta

Where \theta = angle between magnetic field line and surface normal, A = area of circular surface.

A = \pi r^{2}

A = 3.14 \times (0.10) ^{2}

A = 0.0314 m^{2}

Magnetic flux is given by,

\phi = 0.078 \times 0.0314 \times \cos 25

\phi = 2.22 \times 10^{-3} Wb

Therefore, the magnetic flux through surface is 2.22 \times 10^{-3} Wb

6 0
3 years ago
Determine the frequency of a sound wave if it has a speed of 350 m/s and a wavelength of 3.80 m.
Eva8 [605]
Since we have , v=f×lambda (wavelength). Where v equals 350m/s and wavelength equals 3.80. so it will become f = v/lambda=350/3.80=92.1052Hz
7 0
3 years ago
An electron moving to the left at 0.8c collides with a photon moving to the right. After the collision, the electron is moving t
SVETLANKA909090 [29]

Answer:

Wavelength = 2.91 x 10⁻¹² m, Energy = 6.8 x 10⁻¹⁴

Explanation:

In order to show that a free electron can’t completely absorb a photon, the equation for relativistic energy and momentum will be needed, along the equation for the energy and momentum of a photon. The conservation of energy and momentum will also be used.

E = y(u) mc²

Here c is the speed of light in vacuum and y(u) is the Lorentz factor

y(u) = 1/√[1-(u/c)²], where u is the velocity of the particle

The relativistic momentum p of an object of mass m and velocity u is given by

p = y(u)mu

Here y(u) being the Lorentz factor

The energy E of a photon of wavelength λ is

E = hc/λ, where h is the Planck’s constant 6.6 x 10⁻³⁴ J.s and c being the speed of light in vacuum 3 x 108m/s

The momentum p of a photon of wavelenght λ is,

P = h/λ

If the electron is moving, it will start the interaction with some momentum and energy already. Momentum of the electron and photon in the initial and final state is

p(pi) + p(ei) = p(pf) + p(ef), equation 1, where p refers to momentum and the e and p in the brackets refer to proton and electron respectively

The momentum of the photon in the initial state is,

p(pi) = h/λ(i)

The momentum of the electron in the initial state is,

p(ei) = y(i)mu(i)

The momentum of the electron in the final state is

p(ef) = y(f)mu(f)

Since the electron starts off going in the negative direction, that momentum will be negative, along with the photon’s momentum after the collision

Rearranging the equation 1 , we get

p(pi) – p(ei) = -p(pf) +p(ef)

Substitute h/λ(i) for p(pi) , h/λ(f) for p(pf) , y(i)mu(i) for p(ei), y(f)mu(f) for p(ef) in the equation 1 and solve

h/λ(i) – y(i)mu(i) = -h/λ(f) – y(f)mu(f), equation 2

Next write out the energy conservation equation and expand it

E(pi) + E(ei) = E(pf) + E(ei)

Kinetic energy of the electron and photon in the initial state is

E(p) + E(ei) = E(ef), equation 3

The energy of the electron in the initial state is

E(pi) = hc/λ(i)

The energy of the electron in the final state is

E(pf) = hc/λ(f)

Energy of the photon in the initial state is

E(ei) = y(i)mc2, where y(i) is the frequency of the photon int the initial state

Energy of the electron in the final state is

E(ef) = y(f)mc2

Substitute hc/λ(i) for E(pi), hc/λ(f) for E(pf), y(i)mc² for E(ei) and y(f)mc² for E(ef) in equation 3

Hc/λ(i) + y(i)mc² = hc/λ(f) + y(f)mc², equation 4

Solve the equation for h/λ(f)

h/λ(i) + y(i)mc = h/λ(f) + y(f)mc

h/λ(f) = h/lmda(i) + (y(i) – y(f)c)m

Substitute h/λ(i) + (y(i) – y(f)c)m for h/λ(f)  in equation 2 and solve

h/λ(i) -y(i)mu(i) = -h/λ(f) + y(f)mu(f)

h/λ(i) -y(i)mu(i) = -h/λ(i) + (y(f) – y(i))mc + y(f)mu(f)

Rearrange to get all λ(i) terms on one side, we get

2h/λ(i) = m[y(i)u(i) +y(f)u(f) + (y(f) – y(i)c)]

λ(i) = 2h/[m{y(i)u(i) + y(f)u(f) + (y(f) – y(i))c}]

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

Calculate the Lorentz factor using u(i) = 0.8c for y(i) and u(i) = 0.6c for y(f)

y(i) = 1/[√[1 – (0.8c/c)²] = 5/3

y(f) = 1/√[1 – (0.6c/c)²] = 1.25

Substitute 6.63 x 10⁻³⁴ J.s for h, 0.511eV/c2 = 9.11 x 10⁻³¹ kg for m, 5/3 for y(i), 0.8c for u(i), 1.25 for y(f), 0.6c for u(f), and 3 x 10⁸ m/s for c in the equation derived for λ(i)

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

λ(i) = 2(6.63 x 10-34)/[(9.11 x 10-31)(3 x 108){(5/3)(0.8) + (1.25)(0.6) + ((1.25) – (5/3))}]

λ(i) = 2.91 x 10⁻¹² m

So, the initial wavelength of the photon was 2.91 x 10-12 m

Energy of the incoming photon is

E(pi) = hc/λ(i)

E(pi) = (6.63 x 10⁻³⁴)(3 x 10⁸)/(2.911 x 10⁻¹²) = 6.833 x 10⁻¹⁴ = 6.8 x 10⁻¹⁴

So the energy of the photon is 6.8 x 10⁻¹⁴ J

6 0
3 years ago
When magma cools quickly, what kind of texture or
Ksivusya [100]

Answer:

a

Explanation:

when magma cools Crystal's form because the solution is super saturated with respect to some minerals if the magma cools quickly the crystals do not have much time to form hence they are small and also the resulting rock is fine grained

6 0
3 years ago
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