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SSSSS [86.1K]
3 years ago
11

Hi.

Physics
1 answer:
docker41 [41]3 years ago
7 0
<h3>Answer :</h3>

Let the final temperature be "T".

For the piece of copper :

  • mass, \sf{m_c=40\ g.}

  • specific heat capacity, \sf{c_c=0.4\ J\,g^{-1}\,K^{-1}.}

  • initial temperature, \sf{T_c=200^{\circ}C.}

Then the heat of copper :

\sf{\dashrightarrow Q_c=m_cc_c\,\Delta\!T_c}

\sf{\dashrightarrow Q_c =16(T-200)\ J}

For copper calorimeter :

  • mass, \sf{m_{cc} =60\ g.}

  • specific heat capacity, \sf{c_{cc} =0.4\ J\,g^{-1}\,K^{-1}.}

  • initial temperature, \sf{T_{cc} =25^{\circ}C.}

Then the heat of copper calorimeter :

\sf{\dashrightarrow Q_{cc} =m_{cc}c_{cc}\,\Delta\!T_{cc}}

\sf{\dashrightarrow Q_{cc} =24(T-25)\ J}

For water :

  • mass, \sf{m_w=50\ g. }

  • specific heat capacity, \sf{c_w= 4.2\ J\,g^{-1}\,K^{-1}.}

  • initial temperature, \sf{T_w=25^{\circ}C.}

Then heat of water :

\sf{\dashrightarrow Q_w=m_wc_w\,\Delta\!T_w}

\sf{\dashrightarrow Q_w=210(T-25)\ J}

By energy conservation, the sum of all these energies should be zero as there were no heat energy change before the process, i.e.,

\sf{\dashrightarrow Q_c+Q_{cc}+Q_w=0}

\sf{\dashrightarrow16(T-200)+24(T-25)+210(T-25)=0}

\sf{\dashrightarrow 250T- 9050=0}

\sf{\dashrightarrow T=36.2^{\circ}C}

\large \underline{\underline{\boxed{\sf T=36.2^{\circ}C}}}

<u>____________________________</u>

[Note: in case of considering temperature difference it's not required to convert the temperatures from \sf{^{\circ}C} to K or K to \sf{^{\circ}C}.]

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6.3\ m/s^2

Explanation:

Given that,

If the mass of a body is 15 kg and produced an 4.2 m/s², we need to find the acceleration if the mass is 10 kg and the same force is applied.

Force is given by :

F = ma

Since force is same

m_1a_1=m_2a_2\\\\a_2=\dfrac{m_1a_1}{m_2}\\\\a_2=\dfrac{15\times 4.2}{10}\\\\a=6.3\ m/s^2

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8 0
3 years ago
What happens to a light ray if it is incident on a reflective surface along the normal?
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Answer: The incident ray and the reflected ray and the normal will be parallel to each other.

Explanation:

The normal is perpendicular to the surface of the mirror or the reflective surface.

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If a light ray is incident on a reflective surface along the normal. The angle of incidence will be at 90 degrees which will be perpendicular to the surface of the mirror, the reflected ray will bounce back likewise at the same angle which will be perpendicular to the reflective surface.

Both the incident ray and the reflected ray and the normal will be parallel to each other.

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A box of mass 12 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.42. What is th
vladimir2022 [97]

As we know that friction force on box is given by

F_s = \mu_s N

here we know that

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here we have

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4 years ago
Read 2 more answers
A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk
Murrr4er [49]

Answer:

\omega_2=5.1rad/s

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L_1=L_2

Or:

I_1\omega_1=I_2\omega_2

And we want to calculate:

\omega_2=\frac{I_1\omega_1}{I_2}

The total moment of inertia will be the sum of the moment of intertia of the disk of mass m_D=119.1 kg and radius r_D=3.23m, which is I_D=\frac{m_Dr_D^2}{2}, and the moment of intertia of the student of mass m_S=54.3kg at position r (which will be r_1=r=3.23m or r_2=1.39m) will be I_{S}=m_Sr_S^2, so we will have:

\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}

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\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}

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3 years ago
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Law Incorporation [45]

Answer:

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a_c = \frac{v^2}{r}

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Therefore, the speed of the particle is 2.86 m/s

6 0
4 years ago
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