This is a simple problem where you would use the simplified formula
P = m • v
16 kgm/s = 8 kg • v
V = 2 m/s.
The ball is travelling at 2 m/s.
Umm. False i guess I don’t really know
Via the half-life equation we have:
Where A is the amount initially and final amount after some time has passed, t is the time elapsed, h is the half life time and so:
Therefore there will be 2.8125 grams left after 140 seconds.
Answer: The volume of the balloon at the center of the typhoon is 41.7L.
Note: The complete question is given below;
If a small weather balloon with a volume of 40.0 L at a pressure of 1.00 atmosphere was deployed at the edge of Typhoon Odessa, what was the volume of the balloon when it reached the center?
The severity of a tropical storm is related to the depressed atmospheric pressure at its center. In August 1985, Typhoon Odessa in the Pacific Ocean featured maximum winds of about 90 mi/hr and pressure that was 40.0 mbar lower at the center than normal atmospheric pressure. In contrast, the central pressure of Hurricane Andrew (pictured) was 90.0 mbar lower than its surroundings when it hit south Florida with winds as high as 165 mi/hr.
Explanation:
Since no temperature changes were given, it is assumed to be constant. Therefore, Boyle's law which describes the relationship between pressure and volume is used to determine the new volume at the center of Typhoon Odessa. Mathematically, Boyle's law states that; P1V1 = P2V2
Assuming 1atm = 1 bar, 1mbar = 0.001atm, 40mbar = 0.040atm
P1 = 1.0atm, V1 = 40.0L, P2 = 1atm - 0.040atm = 0.960atm, V2 = ?
Using P1V1 = P2V2
V2 = P1V1/P2
V2 = 1.0 * 40.0 / 0.96
V2 = 41.67L
Therefore, the volume of the balloon at the center of the typhoon is 41.7L.
