it would be at least 9.8m/s
Calculate the magnitude of the linear momen- tum for each of the following cases a) a proton with mass 1.67 × 10-27 kg mov- ing with a velocity of 6 × 106 m/s. Answer in units of kg · m/s.
Answer:
7.0 s, 69 m/s
Explanation:
If we take down to be positive, then the time to reach the ground is:
x = x₀ + v₀ t + ½ at²
240 m = (0 m) + (0 m/s) t + ½ (9.8 m/s²) t²
t = 7.0 seconds
The final velocity is:
v² = v₀² + 2a(x - x₀)
v² = (0 m/s)² + 2(9.8 m/s²) (240 m - 0 m)
v = 69 m/s
Answer:
8.8 × 10⁻³ g/L
Explanation:
NaF is a strong electrolyte that ionizes according to the following reaction.
NaF(aq) → Na⁺(aq) + F⁻(aq)
Then, the concentration of F⁻ will also be 0.10 M.
In order to find the solubility of PbF₂ (S), we will use an ICE Chart.
PbF₂(s) ⇄ Pb²⁺(aq) + 2 F⁻(aq)
I 0 0.10
C +S +2S
E S 0.10 + 2S
The solubility product (Kps) is:
Kps = 3.6 × 10⁻⁸ = [Pb²⁺].[F⁻]² = S . (0.10 + 2S)²
In the term 0.10 + 2S, 2S is negligible in comparison with 0.10 and we can omit it to simplify calculations.
Kps = 3.6 × 10⁻⁸ = S . (0.10)²
S = 3.6 × 10⁻⁵ M
The molar mass of PbF₂ is 245.20 g/mol. The solubility of PbF₂ in g/L is:
3.6 × 10⁻⁵ mol/L × 245.20 g/mol = 8.8 × 10⁻³ g/L
The force is greatest when the wire is perpendicular to the magnetic field.
In fact, the expression for the magnetic force is
where I is the current, L the length of the wire, B the magnetic field intensity and
is the angle between the directions of I and B.
We can see that F is maximum when
, so when
, so when the wire and the magnetic field are perpendicular to each other.
