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Talja [164]
2 years ago
13

Find an expression for the minimum frictional coefficient needed to keep a car with speed v on a banked turn of radius R designe

d for speed v0.
Physics
1 answer:
solniwko [45]2 years ago
4 0
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr 
</span><span>Θ = arctan(v0² / gr) </span>

<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>

<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
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romanna [79]

Answer:

A) F=-20.16×10⁹N

B) if the distance doubles, force is 4 times smaller.

Explanation:

q1=-28C

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d=25cm=0.25m

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F=9×10⁹×(-28)×0.005/0.25²

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The minus sign indicates attraction.

If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.

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3 years ago
We need to find the launch velocity of our new marble launcher. we know that it will launch a 25g marble to a distance of 73 cm,
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The launch velocity of the marble launcher is 34.65 m/s

Given that the launch velocity of marble launcher, launches a 25g marble to a distance of 73 cm (0.73 m) and the marble roll up to 6.2 meters before stopping. The launch height is 20 cm (0.2 m).

The time for landing can be calculated by the second equation of motion formula:

h = ut + \frac{1}{2}gt^{2}

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Now, the launch velocity of the marble launcher can be calculated by:

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Speed = \frac{6.93}{0.2}

Speed = 34.65 m/s

Therefore, the launch velocity of the marble launcher is 34.65 m/s

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11 months ago
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Salsk061 [2.6K]

Answer:

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AnnyKZ [126]

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