A mercury atom in the ground state absorbs 20.00 electronvolts of energy and is ionized by losing an electron. How much kinetic
1 answer:
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Hypothesis: An increase in voltage should result in a corresponding rise in current because according to Ohm's Law,
Ohm's Law indicates that current is proportional to voltage when resistance remains constant. Hence, if resistance stays the same, elevating the voltage will lead to an increase in current. Conversely, if voltage remains unchanged and resistance increases, current will decrease.
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.
Answer:
F = 0.535 N
Explanation:
We will apply energy concepts, considering both the peak and the bottom of the path.
Top
Em₀ = U = mg y
Bottom
= K = ½ m v²
Emo =
mg y = ½ m v²
v = √ (2gy)
y = L - L cos θ
v = √ (2g L (1 - cos θ))
Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.
F = ma
a = v² / r
For the turning radius, the cable length is r = L.
F = m 2g (1 - cos θ)
Now, let's find the result.
F = 2 1.25 9.8 (1 - cos 12)
F = 0.535 N
Answer:
a
The density of volume charge is
b
The density of surface charge is
Explanation:
The question states that
The radius measures R
The length is L
The speed is v
The ion count per unit volume is n
The charge per ion is q
The surface thickness of the cylinder is
The volume charge density is mathematically expressed as
The surface charge density is mathematically expressed as
substituting for