One side drops into fault
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.
Answer:
3.35 atm
Since P₂ > 3.00 atm, the lighter would explode.
Explanation:
Step 1: Given data
- Initial pressure of butane gas (P₁): 2.50 atm
- Initial temperature of butane gas (T₁): 293 K
- Final pressure of butane gas (P₂): ?
- Final temperature of butane gas (T₂): 393 K
Step 2: Calculate the final pressure of butane gas
If we assume ideal behavior, we can calculate the final pressure of butane gas using Gay Lussac's law.
P₁/T₁ = P₂/T₂
P₂ = P₁ × T₂/T₁
P₂ = 2.50 atm × 393 K/293 K = 3.35 atm
Since P₂ > 3.00 atm, the lighter would explode.
Answer:
C- 250+250= 500 newtons (the amount the both exerted) :)
Explanation:
Answer: Holmium-165 = Er-165 + electron
Explanation:
Beta decay is the decay of a neutron into a proton and electron. This will increase the number of protons in the Holmium nucleus by one, making it Er, Erbium, with the same atomic mass.
Holmium-165 = Er-165 + electron
The mass is essentially the same, as the proton is assigned the same mass as the neutron it came from.
