The volume of the balloon will halve
Explanation:
Boyle's law states that for an ideal gas kept at constant temperature, the pressure of the gas is proportional to its volume. Mathematically,
![pV=const.](https://tex.z-dn.net/?f=pV%3Dconst.)
where
p is the gas pressure
V is the volume
The equation can also be rewritten as
![p_1 V_1 = p_2 V_2](https://tex.z-dn.net/?f=p_1%20V_1%20%3D%20p_2%20V_2)
And if we apply it to the gas inside the balloon in this problem (assuming its temperature is constant), we have:
is the initial pressure at sea level (the atmospheric pressure)
is the initial volume
is the final pressure
is the final volume
Substituting into the equation, we find:
![V_2 = \frac{p_1 V_1}{p_2}=\frac{(101)V_1}{202}=\frac{V_1}{2}](https://tex.z-dn.net/?f=V_2%20%3D%20%5Cfrac%7Bp_1%20V_1%7D%7Bp_2%7D%3D%5Cfrac%7B%28101%29V_1%7D%7B202%7D%3D%5Cfrac%7BV_1%7D%7B2%7D)
Which means that the volume of the balloon will halve.
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The speed of the block when the compression is 15 cm is 9.85 m/s.
The given parameters;
- <em>mass of the block, m = 2.4 kg</em>
- <em>height of the block, h = 5 m</em>
- <em>compression of the spring, x = 25 cm = 0.25 m</em>
The spring constant is calculated as follows;
![F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{2.4 \times 9.8}{0.25} \\\\k = 94.08 \ N/m](https://tex.z-dn.net/?f=F%20%3D%20kx%5C%5C%5C%5Cmg%20%3D%20kx%5C%5C%5C%5Ck%20%3D%20%5Cfrac%7Bmg%7D%7Bx%7D%20%5C%5C%5C%5Ck%20%3D%20%5Cfrac%7B2.4%20%5Ctimes%209.8%7D%7B0.25%7D%20%5C%5C%5C%5Ck%20%3D%2094.08%20%5C%20N%2Fm)
The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;
![\Delta K.E = \Delta P.E\\\\\frac{1}{2} m(v^2 - v_{0 }^2 ) = mgh - \frac{1}{2} kx^2\\\\\frac{1}{2} mv^2 = mgh - \frac{1}{2} kx^2\\\\mv^2 = 2mgh - kx^2\\\\v^2 = \frac{2mgh - kx^2}{m} \\\\v = \sqrt{\frac{2mgh - kx^2}{m}} \\\\v = \sqrt{\frac{(2 \times 2.4 \times 9.8 \times 5) - (94.08 \times 0.15^2)}{2.4}} \\\\v = 9.85 \ m/s](https://tex.z-dn.net/?f=%5CDelta%20K.E%20%3D%20%5CDelta%20P.E%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20m%28v%5E2%20%20-%20v_%7B0%20%7D%5E2%20%29%20%3D%20mgh%20-%20%5Cfrac%7B1%7D%7B2%7D%20kx%5E2%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%20%3D%20mgh%20-%20%20%20%5Cfrac%7B1%7D%7B2%7D%20kx%5E2%5C%5C%5C%5Cmv%5E2%20%20%20%3D%202mgh%20-%20kx%5E2%5C%5C%5C%5Cv%5E2%20%3D%20%5Cfrac%7B2mgh%20-%20kx%5E2%7D%7Bm%7D%20%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B2mgh%20-%20kx%5E2%7D%7Bm%7D%7D%20%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B%282%20%5Ctimes%202.4%20%5Ctimes%209.8%20%5Ctimes%205%29%20-%20%2894.08%20%5Ctimes%200.15%5E2%29%7D%7B2.4%7D%7D%20%5C%5C%5C%5Cv%20%3D%209.85%20%5C%20m%2Fs)
Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.
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