Answer:
![V_2 = 606.9 m^3](https://tex.z-dn.net/?f=V_2%20%3D%20606.9%20m%5E3)
Explanation:
By ideal gas equation law we know that
PV = nRT
now we know that when balloon rises to certain level then the number of moles will remains same
so we can say
![n_1 = n_2](https://tex.z-dn.net/?f=n_1%20%3D%20n_2)
![\frac{P_1V_1}{RT_1} = \frac{P_2V_2}{RT_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BRT_1%7D%20%3D%20%5Cfrac%7BP_2V_2%7D%7BRT_2%7D)
now plug in all data to find the final volume of the balloon
![\frac{754\times 219}{R(298)} = \frac{210 \times V_2}{R(230)}](https://tex.z-dn.net/?f=%5Cfrac%7B754%5Ctimes%20219%7D%7BR%28298%29%7D%20%3D%20%5Cfrac%7B210%20%5Ctimes%20V_2%7D%7BR%28230%29%7D)
![V_2 = \frac{230\times 754 \times 219}{210 \times 298}](https://tex.z-dn.net/?f=V_2%20%3D%20%5Cfrac%7B230%5Ctimes%20754%20%5Ctimes%20219%7D%7B210%20%5Ctimes%20298%7D)
![V_2 = 606.9 m^3](https://tex.z-dn.net/?f=V_2%20%3D%20606.9%20m%5E3)
Answer:
v = 5.42 m/s
Explanation:
given,
height of the jumper = 1.5 m
velocity of sprinter = ?
kinetic energy can be transformed into potential energy
![m g h = \dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=m%20g%20h%20%3D%20%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
![g h = \dfrac{1}{2}v^2](https://tex.z-dn.net/?f=g%20h%20%3D%20%5Cdfrac%7B1%7D%7B2%7Dv%5E2)
![v =\sqrt{2gh}](https://tex.z-dn.net/?f=v%20%3D%5Csqrt%7B2gh%7D)
![v =\sqrt{2\times 9.8 \times 1.5}](https://tex.z-dn.net/?f=v%20%3D%5Csqrt%7B2%5Ctimes%209.8%20%5Ctimes%201.5%7D)
v = 5.42 m/s
Speed of the sprinter is equal to v = 5.42 m/s
The mass would be 51 kg. So b is the right answer.