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kari74 [83]
2 years ago
7

A van of mass 5,000kg travelling at a velocity of 10 m/s collides with a stationary car. The vehicles move together after the im

pact at a velocity of 6 m/s. Calculate the mass of the car.
Physics
1 answer:
Lunna [17]2 years ago
5 0

Hi there!

We can use the conservation of momentum for an INELASTIC collision:

m_1v_1 + m_2v_2 = v_f(m_1 + m_2)

Let:

m1 = mass of van (5000 kg)

m2 = mass of car (? kg)

v1 = initial velocity of van (10 m/s)

v2 = initial velocity of car (0 m/s)

vf = final velocity of BOTH objects (6 m/s)

We can rearrange and plug in values to solve for m2:

m_1v_1 + m_2(0) = v_f(m_1 + m_2)\\\\m_1v_1 = v_f(m_1 + m_2)\\\\(5000)(10) = 6(5000 + m_2)

Solve:

50000 = 30000 + 6m_2\\\\20000 = 6m_2\\\\m_2 = 20000/6 = \boxed{3333.33 kg}

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A solid object has a mass of 104 kg and a volume of 1,278 m3. What is its density?
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What is the speed of a wave that has a frequency of 200 Hz and a
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25.5 m/s in 5.75 s. What is the acceleration?
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3 years ago
A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

\omega = 2\pi f

Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.2320W

PART B) The new amplitude A' that is half ot the wavelength of the wave is

A' = \frac{1.8*10^{-3}}{2}

A' = 0.9*10^{-3}

Replacing at the equation of power we have that

P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

8 0
3 years ago
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