Answer:
Explanation:
The data indicates that point A is located midway between two charges.
To calculate the electric field at point A, we begin with the field produced by charge -Q ( 6e⁻ ) at A:
= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴
= 13.82 x 10⁻⁶ N/C
This field points towards Q⁻.
A similar field will arise from the charge Q⁺, but it will direct away from Q⁺ toward Q⁻.
To find the resultant field, we add these contributions:
= 2 x 13.82 x 10⁻⁶
= 27.64 x 10⁻⁶ N/C
For the force acting on an electron placed at A:
= charge x field
= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶
= 44.22 x 10⁻²⁵ N
Let's consider a few possibilities.
1. The lowest velocity of the paratrooper would be just before hitting the ground.
2. Given that the jump originated from a relatively short height, the paratrooper utilized a static line, allowing the parachute to deploy almost instantly after leaping.
Hence, we will convert 100 mi/h to ft/s:
100 mi/h * 5280 ft/mi / 3600 s/h = 146.67 ft/sec.
Based on the first assumption, the maximum distance fallen by the paratrooper would equate to 8 seconds at 146.67 ft/s, translating to
8 s * 146.67 ft/s = 1173.36 ft.
This calculated distance is nearly on par with the jump height, validating both assumptions 1 and 2. Thus, this scenario seems plausible.
Moreover, considering the terminal velocity for a parachutist in a freefall position with limbs spread out typically reaches 120 mi/h, which is slightly above the 100 mi/h mentioned in the article. This as well aligns with the notion of the parachute acting like a flag, adding some air resistance.
Answer:
The molar mass of the metal in grams per mole is calculated to be 8.87.
Explanation:
Initially, we can consider a sample of the compound weighing 100 g. This results in:
- 52.92% metal: 52.92 g M
- 47.80% oxygen: 47.80 g O
By utilizing the molar mass of oxygen, which is 16 g / mol, we can determine the quantity of moles of oxygen in the sample via the rule of three:
moles of oxygen=2.9875
The formula for the metal oxide indicates that:
2 M⁺³ + 3 O²⁻ ⇒ M₂O₃
From the previous equation, it is evident that 3 oxygen ions are necessary to react with 2 metal ions. Hence:
Given 52.92 g of metal in the sample, the molar mass of the metal is:
molar mass≅ 8.87 g/mol
The molar mass of the metal in grams per mole is 8.87.
The value that most closely corresponds to this is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.
