The answers would be:
C. Bubbles appear
E. The color changes
<u>If you'd like to know more, you can read on:</u>
Although your selections are indications of chemical reactions, if you are referring to only the chemical reaction that occurred in your example, then the two answers are the only choices.
A chemical reaction occurs when the reactants form a new product. Signs of a chemical reaction include change is smell, color and temperature. In some cases, precipitate forms.
At the same speed as the other ice cube
As the skateboard rolls down the ramp it loses potential energy and gains kinetic energy.
here given that the velocity of the probe is
![v_x = - v sin\omega t](https://tex.z-dn.net/?f=v_x%20%3D%20-%20v%20sin%5Comega%20t)
![v_y = v cos\omega t](https://tex.z-dn.net/?f=v_y%20%3D%20v%20cos%5Comega%20t)
now at initial position where t = 0
![v_{xi} = - v sin0 = 0](https://tex.z-dn.net/?f=v_%7Bxi%7D%20%3D%20-%20v%20sin0%20%3D%200)
![v_{yi} = v cos 0 = v](https://tex.z-dn.net/?f=v_%7Byi%7D%20%3D%20v%20cos%200%20%3D%20v)
Now after t = 24 minutes we need to find final components of velocity
![v_{xf} = - v sin(1.20* 10^{-3} * 24*60) = -v sin(1.728)](https://tex.z-dn.net/?f=v_%7Bxf%7D%20%3D%20-%20v%20sin%281.20%2A%2010%5E%7B-3%7D%20%2A%2024%2A60%29%20%3D%20-v%20sin%281.728%29)
![v_{yf} = v cos(1.20* 10^{-3} * 24*60) = v cos(1.728)](https://tex.z-dn.net/?f=v_%7Byf%7D%20%3D%20%20v%20cos%281.20%2A%2010%5E%7B-3%7D%20%2A%2024%2A60%29%20%3D%20v%20cos%281.728%29)
now as we know that acceleration is given as
![a = \frac{v_f - v_i}{t}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bv_f%20-%20v_i%7D%7Bt%7D)
Now for x direction of motion
![a_x = \frac{v_{xf} - v_{xi}}{t}](https://tex.z-dn.net/?f=a_x%20%3D%20%5Cfrac%7Bv_%7Bxf%7D%20-%20v_%7Bxi%7D%7D%7Bt%7D)
![a_x = \frac{- v sin(1.728) - 0}{24*60}](https://tex.z-dn.net/?f=a_x%20%3D%20%5Cfrac%7B-%20v%20sin%281.728%29%20-%200%7D%7B24%2A60%7D)
![a_x = \frac{-7.77*10^3 * 0.99}{24*60}](https://tex.z-dn.net/?f=a_x%20%3D%20%5Cfrac%7B-7.77%2A10%5E3%20%2A%200.99%7D%7B24%2A60%7D)
![a_x = -5.33 m/s^2](https://tex.z-dn.net/?f=a_x%20%3D%20-5.33%20m%2Fs%5E2)
Now for y direction of motion
![a_y = \frac{v_{yf} - v_{yi}}{t}](https://tex.z-dn.net/?f=a_y%20%3D%20%5Cfrac%7Bv_%7Byf%7D%20-%20v_%7Byi%7D%7D%7Bt%7D)
![a_y = \frac{ v cos(1.728) - v}{24*60}](https://tex.z-dn.net/?f=a_y%20%3D%20%5Cfrac%7B%20v%20cos%281.728%29%20-%20v%7D%7B24%2A60%7D)
![a_y = \frac{7.77*10^3 * (-0.16) - 7.77 * 10^3}{24*60}](https://tex.z-dn.net/?f=a_y%20%3D%20%5Cfrac%7B7.77%2A10%5E3%20%2A%20%28-0.16%29%20-%207.77%20%2A%2010%5E3%7D%7B24%2A60%7D)
![a_y = -6.24 m/s^2](https://tex.z-dn.net/?f=a_y%20%3D%20-6.24%20m%2Fs%5E2)
now in order to find the magnitude of acceleration we can say
![a = \sqrt{a_x^2 + a_y^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Csqrt%7Ba_x%5E2%20%2B%20a_y%5E2%7D)
![a = \sqrt{5.33^2 + 6.24^2} = 8.2 m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%5Csqrt%7B5.33%5E2%20%2B%206.24%5E2%7D%20%3D%208.2%20m%2Fs%5E2)