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Vesnalui [34]
3 years ago
13

An 85.0-kg mountain climber plans to swing down, starting from rest, from a ledge using a light rope 6.50 m long. he holds one e

nd of the rope, and the other end is tied higher up on a rock face. since the ledge is not very far from the rock face, the rope makes a small angle with the vertical. at the lowest point of his swing, he plans to let go and drop a short distance to the ground.
Physics
1 answer:
Rainbow [258]3 years ago
4 0

Understanding the given:
85 kg mountain climber
6.50 m long rope
gravity = 10m/s2

If we want to identify the work done on this scenario 
we get f = 85kg x 10m/s2 = 850 N
w = 850N x 6.5 m = 5525 J

Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
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A converging-diverging nozzle has a throat area of 10 cm2 and an exit area of 28.96 cm2 . A normal shock stands in the exit when
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The tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

The given parameters:

  • <em>Throat area of the nozzle, </em>A^*<em> = 10 cm² = 0.001 m²</em>
  • <em>The exit area of the nozzle, A = 28.96 cm² = 0.002896 m²</em>
  • <em>Air pressure at sea level = 101.325 kPa</em>

The ratio of the areas of the converging-diverging nozzle is calculated as follows;

= \frac{A}{A^*} \\\\= \frac{0.002896}{0.001} \\\\= 2.896

From supersonic isentropic table, at \frac{A}{A^*} = 2.896, we can determine the following;

M_e = 2.6 \ kg/s\\\\\frac{P_o}{P_e} = 19.954

The tank pressure is calculated as follows;

\frac{P_o}{P_e} = 19.954 \\\\P_e = \frac{P_o}{19.954} \\\\P_e = \frac{101.325 \ kPa}{19.954} \\\\P_e = 5.08 \ kPa

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Learn  more about converging-diverging nozzle design here: brainly.com/question/13889483

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