Answer:
The turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1.
Explanation:
Given:
The concentration of bovine carbonic anhydrase = total enzyme concentration = Et = 3.3 pmol⋅L^–1 = 3.3 × 10^–12 mol.L^–1
The maximum rate of reaction = Rmax (Vmax) = 222 μmol⋅L^–1⋅s^–1 = 222 × 10^–6 mol.L^–1⋅s^–1
The formula for the turnover number of an enzyme (kcat, or catalytic rate constant) = Rmax ÷ Et = 222 × 10^–6 mol.L^–1⋅s^–1 ÷ 3.3 × 10^–12 mol.L^–1 = 67,272,727.27 s^–1
Therefore, the turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1
Show transcribed image text. Compare the work done by the electric field when the particle travels<span> from point W to point X to that </span>done<span> when the </span>particle travels<span> from point Z to point Y. ... Is the </span>work done<span> on the </span>particle<span> by the </span>electric field<span> positive, negative, or zero? Explain using force and displacement vectors.</span>
Answer:
1. 20.54m/s
2. 1.52s
Explanation:
QUESTION 1:
The speed the stone impact the ground is the final speed/velocity, which can be calculated using the formula:
v² = u² + 2as
Where;
v = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration due to gravity (m/s²)
s = distance (m)
From the provided information, u = 5.65m/s, v = ?, s = 19.9m, a = 9.8m/s²
v² = 5.65² + 2 (9.8 × 19.9)
v² = 31.9225 + 2 (195.02)
v² = 31.9225 + 390.04
v² = 421.9625
v = √421.9625
v = 20.5417
v = 20.54m/s
QUESTION 2:
Using v = u + at
Where v = final velocity (m/s) = 20.54m/s
t = time (s)
u = initial velocity (m/s) = 5.65m/s
a = acceleration due to gravity (m/s²)
v = u + at
20.54 = 5.65 + 9.8t
20.54 - 5.65 = 9.8t
14.89 = 9.8t
t = 14.89/9.8
t = 1.519
t = 1.52s
