An object starts at position 12 on a horizontal line with a reference point of o. What is the position of the object if it moves
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<em>Q</em><em>u</em><em>e</em><em>s</em><em>t</em><em>i</em><em>o</em><em>n</em><em>:</em><em> </em><em>How many km are in 5.6mm? </em>
<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>3 </em>
<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>-6 </em>
<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>-3 </em>
<em>=</em><em>></em><em> </em><em>5.6x10</em><em>^</em><em>6</em>
<em>A</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em>-</em>
<em>E</em><em>x</em><em>p</em><em>l</em><em>a</em><em>n</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em>:</em><em>-</em>
By using the formula-
As 1 with 6 zeros, we convert it into exponential form.
As this above value is fraction type, we can do the reciprocal, thus, the exponent gets a negative value.
Now combine with given question.
Considering the Charles' law, the gas would have a temperature of -109.2 C.
<h3>Charles' law</h3>Finally, Charles' law establishes the relationship between the volume and temperature of a gas sample at constant pressure. This law says that the volume is directly proportional to the temperature of the gas. That is, if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.
Charles' law is expressed mathematically as:
If you want to study two different states, an initial state 1 and a final state 2, the following is true:
In this case, you know:
- P1= 1800 psi
- V1= 10 L
- T1= 20 C= 293 K (being 0 C= 273 K)
- P2= 1800 psi
- V2= 6 L
- T2= ?
You can see that the pressure remains constant, so you can apply Charles's law.
Replacing in the Charles's law:
Solving:
<u><em>T2=163.8 K= -109.2 C</em></u>
The gas would have a temperature of -109.2 C.
Learn more about Charles's law: