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4 years ago

In a machine, work output is less than work input because some energy is converted into thermal energy. true or false.

Physics

2 answers:

tamaranim1 [39]4 years ago

7 0

True ..........................

blondinia [14]4 years ago

4 0

true because im smart like that

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On a speed-time graph, a horizontal line shows the change in speed is ____.

spayn [35]

On a speed/time graph, the height of the line at any point
shows the speed at that moment. If the line is horizontal,
then its height isn't changing, meaning that the speed isn't
changing. It's constant. The change is zero, until the line
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4 0

4 years ago

When a ray of light passes through a boundary between two transparent media, it changes direction. What is this phenomenon known

algol [13]

Answer:

The phenomenon is known as refraction. The correct option is B

Explanation:

Refraction is defined as the property of a change in the direction of light, when it passed from one medium to another. In the process of refraction, the direction to which the rsy of light follows changes as well as the velocity, but the frequency remains the same along the line of propagation.

The extent of propagation is determined by the nature of the medium allowing the propagation.If a ray of light hits the surface of a sheet of glass, some light will be reflected by the surface of the glass. However, much of the light will pass through the glass, because glass is transparent. This effect also o view with mediums such as water.

The reason why the light rays bends away from its normal direction is that light slows down when it passes from the less dense air into the denser glass or water.

Other properties of light includes:

--> light travels in a straight line

--> light can be reflected

--> light can be diffracted

--> light can be polarized

7 0

3 years ago

Hi.

docker41 [41]

<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 $\sf{^{\circ}C}$ .]