Answer:
The answer to your question is:
a) t1 = 2.99 s ≈ 3 s
b) vf = 39.43 m/s
Explanation:
Data
vo = 10 m/s
h = 74 m
g = 9.81 m/s
t = ? time to reach the ground
vf = ? final speed
a) h = vot + (1/2)gt²
74 = 10t + (1/2)9.81t²
4.9t² + 10t -74 = 0 solve by using quadratic formula
t = (-b ± √ (b² -4ac) / 2a
t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)
t = (-10 ± √ 1550.4 ) / 9.81
t1 = (-10 + √ 1550.4 ) / 9.81 t2 = (-10 - √ 1550.4 ) / 9.81
t1 = (-10 ± 39.38 ) / 9.81 t2 = (-10 - 39.38) / 9.81
t1 = 2.99 s ≈ 3 s t2 = is negative then is wrong there are
no negative times.
b) Formula vf = vo + gt
vf = 10 + (9.81)(3)
vf = 10 + 29.43
vf = 39.43 m/s
When they say use energy, you want to use
Total energy = potential energy + kinetic energy = mgh + 1/2mv²
I assume you mean 200 g ball,
so, we know the total distance traveled is going to be 13 - 1.3 = 11.7 m
If the ball just makes it to the top ( 13 m ) , then the ball will stop moving and the kinetic energy will be 0,
therefore, the potential energy at the top will be the total energy of the system = mgh
from this, we say that mgh = 1/2mv² solve for v
<span>
v = sqrt (2gh) = 15.2 m/s </span>
Answer:
The answer is going to be C.
Explanation:
Trust me. Im an expert in physics
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
Answer:
B. Outside the nucleus.
Explanation:
Electrons orbit the nucleus of the atom.
