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Dafna1 [17]
3 years ago
7

The concepts in this problem are similar to those in Multiple-Concept Example 4, except that the force doing the work in this pr

oblem is the tension in the cable. A rescue helicopter lifts a 81.7-kg person straight up by means of a cable. The person has an upward acceleration of 0.729 m/s2 and is lifted from rest through a distance of 7.86 m. (a) What is the tension in the cable
Physics
1 answer:
steposvetlana [31]3 years ago
5 0

Answer:

Explanation:

Given that:

distance (z) = 7.86 m

mass of the person = 81.7 kg

Acceleration (a) = 0.729 m/s²

By using Newton's second law along the vertical axis:

T - mg = ma\\ \\ T = ma + mg \\ \\  T = m(a+g) \\ \\

T = 81.7 (0.729 +9.8)

T = 860.22 N

The work done now is:

W_T = T\times z \\ \\

W_T = 860.22 \times 7.86

\mathbf{W_T =6761.3292\ N}

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Answer:

Explanation:

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we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

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T_2 = 402.36 \ K

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The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

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