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erik [133]
2 years ago
5

A steel cable with Cross Sectional Area 3.00cm² has an elastic limit of 2.40 x 10^8pascals. Find the maximum upward acceleration

that can be given a 1200kg elevated supported by the cable if the stress is not to exceed one-third of the elastic limit.​
Physics
1 answer:
bazaltina [42]2 years ago
3 0

Answer:

Stress = F / A       force per unit area

A = 3.00 cm^2 = 3 E-4  m^2

F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N    max force applied

F/3 = 2.4E4 N  if force not to exceed limit   (= f)

f = M a

a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2      about 2 g

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\qquad\qquad\huge\underline{{\sf Answer}}♨

Heat capacity of body 1 :

\qquad \sf  \dashrightarrow \:m_1s_1

Heat capacity of body 2 :

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it's given that, the the head capacities of both the objects are equal. I.e

\qquad \sf  \dashrightarrow \:m_1s_1 = m_2s_2

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According to given relation :

\qquad \sf  \dashrightarrow \:(m_1 + m_2) s' = m_1s_1 + m_2s_2

\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_1s_1 + m_2s_2}{m_1 + m_2}

\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_2s_2+ m_2s_2}{ \frac{m_2s_2}{s_1} + m_2 }

[ since, m_2s_2 = m_1s_1 ]

\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2m_2s_2}{ m_2(\frac{s_2}{s_1} + 1)}

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\qquad \sf  \dashrightarrow \: s' =  \dfrac{2s_1s_2}{s_1 + s_2}

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