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aniked [119]
3 years ago
9

Dalton’s completing an investigation in the science lab. He observes that a sample of liquid turns to gas at 135°C. What’s this

temperature called? A. boiling point B. freezing point C. melting point D. room temperature E. standard temperature
Physics
2 answers:
FinnZ [79.3K]3 years ago
4 0

Answer:

option a

Explanation:

artcher [175]3 years ago
3 0

The temperature at which the sample of liquid turns to gas at 135 °C is termed as boiling point.

Answer: Option A

<u>Explanation: </u>

The observation of conversion of liquid to gas indicates that there is occurrence of change in the state of matter. The inter-conversion from one state to another can be done by either varying the temperature or by varying the pressure.

In this case, the liquid on heating gets converted to gaseous state after attaining a particular temperature say 135 °C. So, this process of conversion from liquid to gaseous state on heating is termed as boiling.

The temperature at which a liquid converts to gas is termed as the boiling point of that liquid.

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Answer:

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A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res
Natalija [7]

Answer:

The time constant is \tau = 17.5 \ s    

Explanation:

From the question we are told that

   The spring constant is  k = 11.5 \  N/m

   The mass  of the ball is  m_b  = 490 \ g  = 0.49 \ kg

   The amplitude of the  oscillation t the beginning is x =  6.70 cm = 0.067 \  m

    The amplitude after time t is  x_t = 2.20 cm = 0.022 \  m

    The number of oscillation is N  = 30

Generally the time taken to attain the second amplitude is mathematically represented as

       t  = N  *  T                                            Here  T is the period of oscillation

         t = N * 2\pi \sqrt{\frac{m}{k} }

=>     t = 30 * 2 * 3.142 *  \sqrt{\frac{ 0.490}{11.5} }

=>     t = 38.88 \  s

Generally the amplitude at time t is mathematically represented as

         x(t) = x e^{-\frac{at}{2m} }

Here a is the damping  constant so

 at  t = 38.88 \  s ,  x_t = 2.20 cm = 0.022 \  m

So  

     0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }

=>  0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }

taking natural log of both sides

=>  ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }    

=>   a = 0.028

Generally the time constant is mathematically represented as

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=> \tau = 17.5 \ s    

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Zielflug [23.3K]

Answer:F_{v} =\mu_{k} mg

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Explanation:

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coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

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but once crate starts moving the force of friction is reduced  F_{v} =\mu_{k} mg.

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Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as  forces are balanced.

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