Answer:
7.35m/s
Explanation:
Kinetic energy = 1/2 MV^2
From the question
Kinetic energy = 6.8x10^3J
Mass = 225kg
V =?
Using the above formula
We have
6.8x10^3 = 1/2 x 225 x v^2
6.8x10^3 = 112.5 x v^2
Divide both sides by 112.5
We have
6.8x10^3 / 112.5 = v^2
v ^2 = 54.04
V = squared root of 54.04
v = 7.35m/s
Using the formula for permutation, there are 5040 ways that 7 objects are taken 1 at a time.
<h3>What is permutation?</h3>
The permutation of objects refers to the number of ways that an object can be selected from a number of objects.
Given that;
nPr = n!/(n - r)!
Hence;
7P1 = 7!/(7 - 1)!
7P1 = 7 * 6 !/6!
7P1 = 7 * 6 * 5 * 4 * 3 * 2 *1
7P1 = 5040
Learn more about permutation: brainly.com/question/1216161
![\bold{\huge{\underline{ Solution }}}](https://tex.z-dn.net/?f=%5Cbold%7B%5Chuge%7B%5Cunderline%7B%20Solution%20%7D%7D%7D)
<h3><u>Given </u><u>:</u><u>-</u></h3>
- Here , The initial temperature of 3kg of water is 10° C
- The heat supplied to 3kg water is 7,700 J
- The specific heat of water is 4186 J/kg ° C
<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>
- We have to find the final temperature of water .
<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>
Here, we have
- Mass = 3 kg
- Specific heat capacity = 4186 J/kg°C
- Initial temperature = 10° C
- Heat applied = 7,700J
<u>We </u><u>know </u><u>that</u><u>, </u>
- Specific heat is the heat that is required to increase the temperature of 1 kg or unit mass by 1° C or unit ° C
<u>That </u><u>is</u><u>, </u>
![\bold{\red{ C = }}{\bold{\red{\dfrac{ Q}{m{\delta}T}}}}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cred%7B%20C%20%3D%20%7D%7D%7B%5Cbold%7B%5Cred%7B%5Cdfrac%7B%20Q%7D%7Bm%7B%5Cdelta%7DT%7D%7D%7D%7D)
- Here, Change in temperature is ΔT that is, Final temperature - Initial temperature
<u>So</u><u>, </u>
![\sf{{\delta} T = }{\sf{\dfrac{ Q}{Cm}}}](https://tex.z-dn.net/?f=%5Csf%7B%7B%5Cdelta%7D%20T%20%3D%20%7D%7B%5Csf%7B%5Cdfrac%7B%20Q%7D%7BCm%7D%7D%7D)
<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>
![\sf{ T2 - 10 = }{\sf{\dfrac{ 7700}{3{\times}4186}}}](https://tex.z-dn.net/?f=%5Csf%7B%20T2%20-%2010%20%3D%20%7D%7B%5Csf%7B%5Cdfrac%7B%207700%7D%7B3%7B%5Ctimes%7D4186%7D%7D%7D)
![\sf{ T2 - 10 = }{\sf{\dfrac{ 7700}{12558}}}](https://tex.z-dn.net/?f=%5Csf%7B%20T2%20-%2010%20%20%3D%20%7D%7B%5Csf%7B%5Cdfrac%7B%207700%7D%7B12558%7D%7D%7D)
![\sf{ T2 - 10 = }{\sf{\cancel{\dfrac{ 7700}{12558}}}}](https://tex.z-dn.net/?f=%5Csf%7B%20T2%20-%2010%20%20%3D%20%7D%7B%5Csf%7B%5Ccancel%7B%5Cdfrac%7B%207700%7D%7B12558%7D%7D%7D%7D)
![\sf{ T2 - 10 = 0.61 }](https://tex.z-dn.net/?f=%5Csf%7B%20T2%20-%2010%20%3D%200.61%20%7D)
![\sf{ T2 = 0.61 + 10 }](https://tex.z-dn.net/?f=%5Csf%7B%20T2%20%3D%200.61%20%2B%2010%20%7D)
![\bold{ T2 = 10.61 {\degree} C}](https://tex.z-dn.net/?f=%5Cbold%7B%20T2%20%3D%2010.61%20%7B%5Cdegree%7D%20C%7D)
Hence, The final temperature of 3kg if 7,700 J of heat supplied is 10.61 °C