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2 years ago

7

If the car has a mass of 0.4 kg, the ratio of height to width of the ramp is 19/85, the initial displacement is 1.9 m, and the c

hange in momentum is 1.27 kg*m/s, how far will it coast back up the ramp before changing directions (meters)?

1 answer:

Rasek [7]2 years ago

5 0

Answer:

final displacement lf = 0.39 m

Explanation:

from change in momentum equation:

$\delta p = m \sqrt(2g * y/x)* [\sqrt li + \sqrt lf]$

given: m = 0.4kg, y/x = 19/85, li = 1.9 m,

\delta p = 1.27 kg*m/s.

putting all value to get the final displacement value

$1.27 = 0.4\sqrt(2*9.81 *(19/85))* [\sqrt 1.9 + \sqrt lf]$

final displacement lf = 0.39 m

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Answer: 0.258

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$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

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So, in this problem we have two transversal areas:

<u>For aluminium:</u>

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

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<u>For copper:</u>

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

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$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

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$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

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$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

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