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amm1812
3 years ago
8

A long. 1.0 kg rope hangs from a support that breaks, causing the rope to fall, if the pull exceeds 43 N. A student team has bui

lt a 2.0 kg robot "mouse" that runs up and down the rope. What minimum magnitude of the acceleration should the robot have for the rope to fail? Express your answer with the appropriate units.
Physics
1 answer:
raketka [301]3 years ago
8 0

Answer:

6.8 m/s2

Explanation:

Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is

W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N

For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of

a = F/m = 13.6 / 2 = 6.8 m/s2

So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail

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An electron at Earth's surface experiences a gravitational force of magnitude F=(9.11×10−31 kg)⋅(9.8 m/s2). Part A How far away
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Answer:

r = 5,085 m

Explanation:

The force exerted by on the surface of the Earth on an electron is its weight

          W = F = 9.11 10⁻³¹  9.8

          W = 8.9 10⁻³⁰ N

The electric force between an electron and a proton is given by Coulomb's Law

         Fe = k q₁ q₂ / r²

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They ask us that W = Fe

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Let's calculate

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Let's look for the relationship of this distance with the harmonic distance

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