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netineya [11]
3 years ago
12

PLEASE HELP!!!

Physics
2 answers:
nasty-shy [4]3 years ago
4 0

(A) The thermal conductors don't have to be hot to transfer heat, as when the case of ice cube is taken, when the ice cube is placed in the warm water, it melts because the thermal heat from the warm water move into the ice cube, the thermal energy of the ice cube increases whereas the thermal energy of the water decreases.

(B) A heat transfer medium is may be presented in the solid, liquid or the vapor phase. It is used to transfer heat from one place to another or to store heat in a reversible form.

(C) Option A is correct.

This is because, the magnetic field lines transfer from North Pole to South pole which is the correct origination and the destination respectively.

Option B is not correct because the magnetic field lines originate from  North Pole but does not end at the South Pole.

Option C is not correct because the magnetic field lines cannot travel from the same pole to the another same pole of the magnet. In this, the magnetic field lines originate from the South  Pole and ends at the South pole of the other magnet which is not possible.

(D) The law of conservation of energy states that energy can neither be created nor destroyed, it can only be transformed from one form to another.  A heat pump is a device  that can pump thermal energy either into (heating) or out of (cooling) an enclosed space, which means system remains the same but the heat gets converted from one form to the another.


fgiga [73]3 years ago
4 0
The answers are as follows: 
21. A situation where an ice cube will transfer heat to an object that it is in contact with is the is the situation where you put an ice cube inside a warm drink. The ice cube that you put inside the warm drink will melt because, thermal heat from the drink moves into the ice cube; the drink thermal energy decreases while the ice cube thermal energy increases.
22. A medium of heat transfer refers to a substance, which acts as agent in a heat exchanger, through which heat is exchanged from one point to the other. The heat medium may be a liquid or a solid. 
31. The picture is option A is true. This is because, the magnetic field lines start from the correct origin, that is, from the north pole and they end in the south pole.
The picture in option B is not correct because, the magnetic field lines from outside the magnet start form the north pole. The magnetic field lines ought to start from the south pole and end in the north pole, not the other way round.
The picture in option C is incorrect because, magnetic field lines can not travel from a like pole to another like pole. The magnetic lines in this picture travel from the south pole to the south pole, which is not possible, because in physics, like do not attract like.
32. The law of conservation of energy states that energy can neither be created nor destroyed, it can only be transformed from one form to another.
A heat pump transfer heat from a cold area to a warm area. The energy that circulates in the heat pump system remains the same, but it is converted from one form to another form.
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Answer:

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\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

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Given that, the scattering angle is, \theta=147^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

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\lambda^{'}=2.78+4.45=7.23 pm

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\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

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Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

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\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

This is the de Broglie wavelength of the electron after scattering.

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