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marta [7]
3 years ago
12

A middle-A tuning fork vibrates with a frequency f of 440 hertz (cycles per second). You strike a middle-A tuning fork with a fo

rce that produces a maximum pressure of 5 pascals. Write a sine model that gives the pressure P as a function of the time t (in seconds).A middle-A tuning fork vibrates with a frequency f of 440 hertz (cycles per second). You strike a middle-A tuning fork with a force that produces a maximum pressure of 5 pascals. Write a sine model that gives the pressure P as a function of the time t (in seconds).
Physics
1 answer:
jeyben [28]3 years ago
3 0

Answer:

P = 5sin(880πt)

Explanation:

We write the pressure in the form P = Asin2πft where A = amplitude of pressure, f = frequency of vibration and t = time.

Now, striking the middle-A tuning fork with a force that produces a maximum pressure of 5 pascals implies A = 5 Pa.

Also, the frequency of vibration is 440 hertz. So, f = 440Hz

Thus, P = Asin2πft

P = 5sin2π(440)t

P = 5sin(880πt)

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Distance is indeed a scalar amount that also refers to "<em><u>how the ground an object has encased</u></em>", and the Displacement is a vector thing that leads "<em><u>to the extent to which an object is located</u></em>", and the further calculation can be defined as follows:

Given:

distance= 70 miles

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A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
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The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

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Answer:

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Explanation:

Take east to be +x and north to be +y.

The x component of the velocity is:

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The y component of the velocity is:

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vᵧ = 711.4 mph

The speed is found with Pythagorean theorem:

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Rounded to two significant figures, the jet's speed relative to the ground is 1500 mph.

8 0
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