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kherson [118]
3 years ago
14

A coin mass 0.005 kg is dropped from a height of 3 m. How much mechanical energy does it have right before it hits the ground ?

Physics
1 answer:
lesya692 [45]3 years ago
6 0
Mechanical energy equals the sum of potential and kinetic energy. During the process, all PE converts into KE, assuming air resistance is neglected. So, the mechanical energy does not change and is equal to the initial potential energy.
ME
=mgh
=0.005 x 9.81 x 3
=0.147J
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Answer:

The maximum amount of work is  W = 1563.289 \ J

Explanation:

From  the question we are told that

   The temperature of the environment is  T = 280\ K

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    Initially the number of moles  is  n = 1.2 \ moles

     The volume of container B is V_B = 3.5 \ m^3

     

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

             W =  P_A V_A  ln[ \frac{V_B}{V_A} ]

Now from the Ideal gas law

          P_A V_A =  nRT

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Where R is the gas constant with a values of  R =  8.314 \  J/mol

Substituting values we have that

            W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]

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8 0
3 years ago
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3 years ago
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Galina-37 [17]

Answer:

6ms^-1

Explanation:

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2 years ago
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Answer:

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Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

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here we will have

\tau = (m_1g - m_2g)(\frac{L}{2})

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Part b)

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Answer:

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