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3 years ago

Why is it a good idea to start with room temperature water in the calorimeter?

Physics

2 answers:

densk [106]3 years ago

8 0

Explanation:

It is a good idea to start with room temperature water in the calorimeter because the room temperature water helps to determine the heating up/cooling down because of the environment as the experiment takes place. Because the calorimeter heat is the same as the heat of the water.

jeka57 [31]3 years ago

6 0

Answer:

it a good idea to have room temperature water in calorimeter because it has a high specific heat it is non-flammable, it can be measured easily, it is affordable and it can retain the heat for longer time.

Explanation:

For several reasons water at room temperature is used in the calorimeter.

High Specific Heat

It can store high amount of heat with low rise in temperature. It can store heat for longer time which allows the material in the calorimeter to absorb that heat.

High range of Freezing and Boiling point

It has a huge rage between its freezing point and its boiling point and thus we can measure a large range of temperature without its phase change.

Abundance

Water is a pure compound available in abundance with ease.

Definition

According to the definition, one calorie is the amount of energy required by one gram of water to raise its temperature by one degree Celsius. Therefore water plays a vital role in calorie measurement.

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Answer:

1. Hydrogen

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3 years ago

Read 2 more answers

Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec

Dmitrij [34]

Answer:

(a) $F_{net}$ = 4.19 x $10^{-5}$ N, and its direction is towards $m_{2}$ .

(b) It must be placed inside a hollow shell.

Explanation:

Let, $m_{1}$ = 165 kg, $m_{2}$ = 465 kg, $m_{3}$ = 60 kg, and the distance between $m_{1}$ and $m_{2}$ is 0.340 m.

(a) Since $m_{3}$ is placed midway between $m_{1}$ and $m_{2}$ , then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = $\frac{Gm_{1}m_{2} }{r^{2} }$

Where all variables have their usual meaning.

Then,

a. $F_{net}$ = $F_{23}$ - $F_{13}$

$F_{13}$ = $\frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }$

= 2.25 x $10^{-5}$ N

$F_{23}$ = $\frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }$

= 6.44 x $10^{-5}$ N

∴ $F_{net}$ = = 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

The net force exerted by the two masses on the 60 kg object is 4.19 x $10^{-5}$ N.

(ii) / $F_{net}$ / = / $F_{23}$ / - / $F_{13}$ /

= 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

(iii) The direction of the net force is to the right i.e towards $m_{2}$ .

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

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3 years ago

I believe Newton's 1st law is

Bogdan [553]

Answer:

The first law states that if the net force is zero, then the velocity of the object is constant.

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Why is it sometimes difficult for scientists to identify a a fold or fault?

Sauron [17]

Folds and faults are difficult to identify because they occur in the interior of rocks and also due to the dense nature of the materials.

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Faults are lines of weakness are present in materials dues to uneven positioning of the particles of the material.

Folds occurs when infolds occur in materials.

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Sladkaya [172]

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