Answer:
The average speed of the blood in the capillaries is 0.047 cm/s.
Explanation:
Given;
radius of the aorta, r₁ = 1 cm
speed of blood, v₁ = 30 cm/s
Area of the aorta, A₁ = πr₁² = π(1)² = 3.142 cm²
Area of the capillaries, A₂ = 2000 cm²
let the average speed of the blood in the capillaries = v₂
Apply continuity equation to determine the average speed of the blood in the capillaries.
A₁v₁ = A₂v₂
v₂ = (A₁v₁) / (A₂)
v₂ = (3.142 x 30) / (2000)
v₂ = 0.047 cm/s
Therefore, the average speed of the blood in the capillaries is 0.047 cm/s.
Answer:
Let I and j be the unit vector along x and y axis respectively.
Electric field at origin is given by
E= kq1/r1^2 i + kq2/r2^2j
= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)
= (2.88i + 1.44j)*10^-3 N/C
Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3
F= (1.382 i + 0.691 j) *10^-21
Goodluck
Explanation:
Answer:
Work done gets doubled.
Explanation:
The work done by a force is given by :
W = Fd
Where
F is force and d is distance move
If the force is doubled and the distance moved remain the same, it would mean that the work done becomes double of the initial work done.
Answer:
OA. . review all safety procedures and the lab activity procedure
Explanation:
A) It will be 2 covalent bonds
B) covalent bonds occur when there’s 2 atoms that share electrons. In this case by sharing the 2 pairs of valence electrons each atom has a total of 8 valence electrons
