Answer:
Ek1 = 900000 [J]
Ek1 = 400000 [J]
Explanation:
In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:

where:
m = mass = 500 [kg]
v1 = 60 [m/s]
So we have:
Ek1 = 0.5*500*(60^2)
Ek1 = 900000 [J]
and:
Ek2 = 0.5*500*(40^2)
Ek2 = 400000 [J]
D. Electrons are shared between the bromine atoms and carbon atoms
0.74in
Explanation:
Given parameters:
length range of adult mosquito = 3.0 - 6.0mm
Smallest mosquito = 2.5mm
Largest mosquito = 19mm
Average mass range = 3.0 - 5.0mg
Unknown:
Length of largest known mosquito in inches = ?
Solution;
The length is the longest dimension. It is how long a body is.
The problem here is converting from mm to inches;
The length of the longest mosquito which is the largest is 19mm
19mm to inches;
1mm = 0.039inches
19mm = 19mm x
= 0.74in
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Answer:
Torque decreases .
Explanation:
The tape is pulled at constant speed , speed v is constant , so there is
v = ω r where ω is angular speed and r is radius , As radius decreases , angular speed ω increases , So there is angular acceleration .
Let it be α . Let I be moment of inertia of reel .
Reel is in the form of disc
I = 1/2 m r²
I x α = torque
1/2 m r² x α = torque
As the reel is untapped , its mass decreases , r also decreases , so torques also decreases .
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