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KatRina [158]
3 years ago
5

Two tuning forks, 254 Hz. and 260 Hz., are struck simultaneously. How many beats will be heard?

Physics
1 answer:
Lisa [10]3 years ago
7 0

The longer you continue to listen, the more beats will be heard.

They'll occur at the rate of (260Hz - 254Hz) = 6 Hz .
 
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A car is traveling around a horizontal circular track with radius r = 220 m at a constant speed v = 16 m/s as shown. The angle θ
iogann1982 [59]
Velocity = distance / time = ( 2 * pi * r ) / t = 20.583 m/s 

<span>x component = sine ( 32 ° ) * 20.583 = 10.91 m/s

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3 years ago
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boyakko [2]

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3 years ago
In an insulated vessel, a quantity of hot water at temperature T1 is mixed with a different quantity of cold water at temperatur
erastovalidia [21]

Answer:Water Only

Explanation:

Given

vessel is insulated therefore no heat can be added or removed i.e. heat exchange is zero

If hot water at T_1 is mixed with cold water at T_2 then at equilibrium vessel contains only water and final temperature of water will be between T_1 and T_2

Heat released by hot water is equal to heat gain by cold water .

4 0
3 years ago
A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between
elena-14-01-66 [18.8K]

Answer:

-589.05 J

Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

So, substituting the values of the variables into the equation, we have

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

W = -589.047 J

W ≅ -589.05 J

4 0
3 years ago
The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total
pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass m travelling at a velocity \vec{v}. The momentum \vec{p} of this object would be:

\vec{p} = m \cdot \vec{v}.

For the law of conservation of momentum, consider two objects: object \rm a and object \rm b. Assume that these two objects collided with each other.

  • Let m_{\rm a} and m_{\rm b} denote the mass of the two objects.
  • Let \vec{v}_{\rm a}(\text{initial}) and \vec{v}_{\rm b}(\text{initial}) denote the velocity of the two object right before the interaction.
  • Let \vec{v}_{\rm a}(\text{final}) and \vec{v}_{\rm b}(\text{final}) denote the velocity of the two objects right after the interaction.
  • The momentum of the two objects right before the collision would be m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) and m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}), respectively.
  • The momentum of the two objects right after the collision would be m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) and m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final}), respectively.

The sum of the momentum of the two objects would be:

  • m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) right before the collision, and
  • m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final}) right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final}).

4 0
3 years ago
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