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Scorpion4ik [409]
2 years ago
11

An 80-kg firefighter slides down a fire pole. After 1.3 seconds of sliding, the firefighter is sliding at a velocity of 6.5 m/s,

straight down the pole. Once this velocity is reached, the firefighter grips the pole so that the force of friction exerted by the firefighter's hands on the pole is equal to the force of gravity. At this point what is the downward acceleration of the firefighter
Physics
1 answer:
yuradex [85]2 years ago
5 0

Answer:

a= 0

Explanation:

  • In the vertical direction, if the friction force (directed upward) is equal to the force of gravity (downward) this means that no net force is acting on the firefighter.
  • According to Newton's 2nd Law, if no net force is present, the acceleration in this direction is just zero, as follows:

       F_{net} = m*a = 0  (1)

      ⇒ a = 0

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hichkok12 [17]

Answer:

v = 3.5 \times 10^5 m/s

Explanation:

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so we will have

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now we have

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so we will have

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Now by energy conservation

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-6.26 \times 10^{8} - 13046 + \frac{1}{2}v^2 = -1.15 \times 10^6 - 1.28 \times 10^5

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7 0
3 years ago
If the phase angle for a block–spring system in SHM is ϕ and the block's position is given by x = xm cos(ωt + ϕ), what is the ra
matrenka [14]

<h2>K.E/P.E = m/k  tan²φ x ω²</h2>

Explanation:

The given position of block x = x₀ cos(ωt + φ)

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2 years ago
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Answer:

KE = 11,719 J

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