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lidiya [134]
3 years ago
14

A 5-kg block(m1) is released (from rest) from the top of a 3.0-m high, frictionless incline. When the 5-kg block reaches the bot

tom of the incline it collides with a 3-kg block in a perfectly elastic collision. After the collision, the two blocks slide horizontally on a rough surface(μk=0.31) until they come to rest. How far does the 3-kg block slide before coming to rest?
Physics
1 answer:
ollegr [7]3 years ago
3 0

To solve this problem we will apply the concepts related to the conservation of energy (Potential and kinetic), later we can consider the conservation of kinetic energy to finally apply the theory of kinematic equations of linear motion and find the velocity before reaching rest . From the conservation of energy we have to,

\frac{1}{2} mv_1^2 = mgh

v_1 = \sqrt{2gh}

v_1 = \sqrt{2(9.8)(3)}

v_1 = 7.66m/s

From conservation of kinetic energy we have that

KE_i = KE_f

There is not Kinetic Energy at the beginning, then

KE_f = 0

\frac{1}{2} m_2v_2^2 - \frac{1}{2} m_1v_1^2 = 0

\frac{1}{2} m_2v_2^2 = \frac{1}{2} m_1v_1^2

Rearranging to find the velocity 2

v_2 = \sqrt{\frac{m_1}{m_2}}v_1

v_2 = \sqrt{\frac{5}{3}}(7.66)

v_2 = 9.8m/s

From kinetic equation we have that

\Delta V = 2ad

9.899^2 = 2(0.31)(9.8)d

d = 16.12m

Therefore the distance is 16.12m

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In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 1850 J of work is done on the gas
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Answer:

The value of change in internal energy of the gas = + 1850 J

Explanation:

Work done on the gas (W) =  - 1850 J

Negative sign is due to work done on the system.

From the first law  we know that Q = Δ U + W ------------- (1)

Where Q = Heat transfer to the gas

Δ U = Change in internal energy of the gas

W = work done on the gas

Since it is adiabatic compression of the gas so heat transfer to the gas is zero.

⇒ Q = 0

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You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.
Neporo4naja [7]

Answer:

The distance is r_2  =  0.24 \  m

Explanation:

From the question we are told that

       The  distance from the conversation is r_1    =  24.0 \ m

       The  intensity of  the sound at your position is  \beta _1 =  40 dB

        The  intensity at the sound at the new position is  \beta_2 =  80.0dB

Generally the intensity in  decibel is  is mathematically represented as

      \beta  =  10dB log_{10}[\frac{d}{d_o} ]

The intensity is  also mathematically represented as

      d =  \frac{P}{A}

So

    \beta  =  10dB *  log_{10}[\frac{P}{A* d_o} ]

=>   \frac{\beta}{10}  =  log_{10} [\frac{P}{A (l_o)} ]

From the logarithm definition

=>    \frac{P}{A  *  d_o}  =  10^{\frac{\beta}{10} }

=>      P =  A (d_o ) [10^{\frac{\beta }{ 10} } ]

Here P is the power of the sound wave

 and  A is the cross-sectional area of the sound wave  which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

               P_1 =  A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]

Now the power of the sound wave at the second  position is mathematically represented as

               P_2 =  A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]

Generally  power of the wave is constant at both positions  so  

    A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]  = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]

      4 \pi r_1 ^2   [10^{\frac{\beta_1 }{ 10} } ]  = 4 \pi r_2 ^2   [10^{\frac{\beta_2 }{ 10} } ]

        r_2 =  \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}

       substituting value

        r_2 =   \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}

        r_2  =  0.24 \  m

     

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3 years ago
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