The circular lines you see on the chart are isobars, which join areas of the same barometric pressure.
Use Newton’s second law: F=ma
m= 250kg. a= 750ms-2
F= 250 x 750 = 187 500N
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
C)Electric Charges Produce Electric Fields
Given:
Shaft Power, P = 7.46 kW = 7460 W
Speed, N = 1200 rpm
Shearing stress of shaft, 
= 30 MPa
Shearing stress of key, 
= 240 MPa
width of key, w = 
d is shaft diameter
Solution:
Torque, T = 
where,
= 59.365 N-m
Now,
d = 0.0216 m
Now,
w = = 
= 5.4 mm
Now, for shear stress in key
= 
we know that
T = 
= F. 
⇒ = 
⇒ 
= 
length of the rectangular key, l = 4.078 mm
