<span>True
</span><span>True
</span><span>False*
</span><span>False*
</span><span>True
</span><span>True
</span><span>False
A,B,AB,O
10.)?
11.)</span><span>water
carbon dioxide
12.)</span><span>geocentric
</span>13.)<span>Juptier</span>
The best and most correct answer among the choices provided by your question is the fourth choice.
Chlorine and sodium are most likely to form monatomic ions.
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Answer:
The sled slides d=0.155 meters before rest.
Explanation:
m= 60 kg
V= 2 m/s
μ= 0.3
g= 9.8 m/s²
W= m * g
W= 588 N
Fr= μ* W
Fr= 176.4 N
∑F = m * a
a= (W+Fr)/m
a= 12.74m/s²
t= V/a
t= 0.156 s
d= V*t - a*t²/2
d= 0.155 m
Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37
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