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Vsevolod [243]
8 months ago
8

At what temperature will wood and iron apper equally hot or equally cold to our body?

Physics
1 answer:
Nonamiya [84]8 months ago
6 0

They will feel equally hot or cold at the same temperature.

What do you mean by temperature?

Temperature is a physical parameter that describes how hot stuff or radiation is. There are 3 different types of temperature scales: those described in regards to the average translational kinetic energy per freely moving microscopic substance in a body, such as a atom, compound, or anion, such as the SI scale; those that rely solely on purely <u>macroscopic properties and thermodynamic principles</u>, such as Kelvin's original definition; <u>and those that are not defined by theoretical principles, but are defined by convenient empirical properties of particulate matter.</u>  

If they are the same temperature as the area of your body that is touching them, they will feel equally hot or cold. There will be no heat movement if the wood and iron at the same temperature as the skin, and the iron's higher thermal conductivity will be irrelevant. <u>The </u><u>specific temperature</u><u> cannot be specified because if you touch the materials with your fingertips, the wood and iron must match that</u><u> temperature</u><u>, and if you test them with your tongue, which is often considerably </u><u>warmer </u><u>than your hands, the wood and iron must be </u><u>warmer</u><u>.</u>

To know more about temperature, click on the link

brainly.com/question/13165448

#SPJ10

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if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu
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Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

F = \frac{k|q_1||q_2|}{r^2}

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

Therefore, the mutual force between the two point charges is 319.64 N

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Answer:

F_{\text{par}} = F_{\text{frict}}

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