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Anastasy [175]
3 years ago
7

A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end

of the wire, a pulse travels down the wire and returns 7.84 ms later, having reflected off the lower end. The stone is heavy enough to prevent the lower end of the wire from moving. What is the mass m of the stone
Physics
1 answer:
Triss [41]3 years ago
3 0

Answer: m= 35.6 kg

Explanation:

For finding the mass of the stone we have the formula

v= \sqrt{\frac{Tension}{Linear. Mass. density} }

Here, Tension= m*g = m*9.81

and linear mass density= \frac{8.25 g}{65 cm}

Linear mass density= \frac{8.25*10^-3}{65*10^-2}

Linear mass density= 0.0127 kg/m

Velocity= 2*\frac{l}{t}

Velocity= 2 * \frac{65*10^-2}{7.84}

Velocity= 165.8 m/s

So putting all these values in equation we get

v= \sqrt{\frac{Tension}{Linear. Mass. density} }

165.8= \sqrt{\frac{m*9.81}{0.0127} }

Solving we get

m= 35.58 kg

or m= 35.6 kg

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Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

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From the kepler's third law,

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For first satellites,

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{T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

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For second satellites,

   \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} }  }

{T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat2} = 37.67 days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

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