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dmitriy555 [2]
3 years ago
9

A 63.2-kg climber finds herself dangling over the edge of a cliff. Fortunately, she’s connected by a rope of negligible mass to

a 1220-kg rock located 48.6 m from the edge of the cliff. Unfortunately, the ice is frictionless, so the climber accelerates downward. What’s her acceleration, and how much time does she have before the rock goes over the edge?
Physics
1 answer:
wel3 years ago
5 0

Answer:0.535 m/s^2

Explanation:

Given

mass of climber(m_1)=63.2 kg

Distance between rock and cliff=48.6 m

mass of rock(m_2)=1220

let T be the tension in the rope

Thus T-m_1g=m_1a------1

where a is the acceleration of system

Also for Rock

T=m_2a------2

From 1 & 2 we can say that

m_1\left [ g+a\right ]=m_2\left [ a\right ]

g+a=\frac{1220}{63.2}\left [ a\right ]

g+a=19.303 a

g=18.303 a

a=\frac{g}{18.303}

a=0.535 m/s^2

Thus climber is decelerating with 0.535 m/s^2

time to cover 48.6 m

48.6=\frac{0.535t^2}{2}

t=13.47 s

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Statement A  

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Statement B  

Angular acceleration is the rate of change in angular velocity with respect to time.  

Angular velocity and angular acceleration are related by  

{\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t

Which when re-arranged becomes  

\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}

There’s no change in angular velocity anytime when the angular velocity is \omega = 10\,{\rm{rad/s}}.The final and initial velocities remain the same.  

The equation can be modified as follows:  

\begin{array}{c}\\\alpha = \frac{{10\,{\rm{rad/s}} - 10\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}

Therefore, the angular acceleration becomes zero and statement B is valid  

Statement C  

Angular velocity is defined as the change in the angular position with respect to time.  

Angular velocity and angular displacement are related by  

\theta = \omega t

Which can also be modified as:  

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