<u>Acceleration</u> is the rate at which <u>velocity</u> changes.
Answer:
2.5m/s^2
Explanation:
Step one:
given
distance = 20meters
time = 2 seconds
initial velocity u= 0m/s
let us solve for the final velocity
velocity = distance/time
velocity= 20/2
velocity= 10m/s
divide both sides by 40
The purpose of an experiment is to LEARN the EFFECT of something.
The way you do that is to CHANGE the thing and see what happens.
You can change as many things as you want to. But If you change
TWO things and observe the result, then you don't know which one
of them caused the effect you see.
Or maybe BOTH of them working together caused it. You don't know.
So your experiment is not really much good. You need to do it again.
Explanation:
Since its accelerating, the velocity vs time graph is linear
For displacement we need initial velocity (which is zero because it starts from rest) and final velocity (which is calculatee thro acceleration formula
A= (vf - vi)/t
a= vf-0/t
1.25=vf / 7
1.25*7=vf
8.75 = vf
Now for displacement plug all the values in
X = 1/2(vf-vi)/t formula
The displacement (x) is 30.625 m
For part 3, we know new displacement that is 22m , the final and initial velocities are the same so just plug in the values for same formula above
The answer is t = 5.02
Im pretty sure all the answers are correct
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
