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Lynna [10]
2 years ago
14

In this lab, you will use a dynamics track to generate collisions between two carts. If momentum is conserved, what variable cha

nge would result in a velocity change after a collision? In the space below, write a scientific question that you will answer by doing this experiment.
Physics
2 answers:
BartSMP [9]2 years ago
8 0

In collision type of problems since momentum is always conserved

we can say

m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}

So here along with this equation we also required one more equation for the restitution coefficient

v_{2f} - v_{1f} = e(v_{1i} - v_{2i})

so above two equations are required to find the velocity after collision

here the change in velocity occurs due to the contact force while they contact in each other

so this is the impulse of collision while they are in contact with each other while in collision which changes the velocity of two colliding objects

Alborosie2 years ago
5 0

how does changing mass affect colliding objects?

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The total energy (also called mechanical energy) is the sum of the kinetic energy and potential energy:
TE = KE + PE
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1 year ago
A.Whale communication. Blue whales apparently communicate with each other using sound of frequency 17.0 Hz, which can be heard n
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A. 90.1 m

The wavelength of a wave is given by:

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is its frequency

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\lambda=\frac{1531 m/s}{17.0 Hz}=90.1 m

B. 102 kHz

We can re-arrange the same equation used previously to solve for the frequency, f:

f=\frac{v}{\lambda}

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v = 1531 m/s is the wave speed

\lambda=1.50 cm=0.015 m is the wavelength

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f=\frac{1531 m/s}{0.015 m}=1.02 \cdot 10^5 Hz=102 kHz

C. 13.6 m

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\lambda=\frac{v}{f}

where

v = 340 m/s is the speed of sound in air

f = 25.0 Hz is the frequency of the whistle

Substituting into the equation,

\lambda=\frac{340 m/s}{25.0 Hz}=13.6 m

D. 4.4-8.7 m

Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:

- Wavelength corresponding to the minimum frequency (f=39.0 Hz):

\lambda=\frac{340 m/s}{39.0 Hz}=8.7 m

- Wavelength corresponding to the maximum frequency (f=78.0 Hz):

\lambda=\frac{340 m/s}{78.0 Hz}=4.4 m

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E. 6.2 MHz

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\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4} m

And since the speed of the sound wave is

v = 1550 m/s

The frequency will be

f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz

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Answer:

E=0

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