Answer:
Charge enter a 0.100 mm length of the axon is 
Explanation:
Electric field E at a point due to a point charge is given by
where 
is the constant =
is the magnitude of point charge and 
is the distance from the point charge
Charges entering one meter of axon is 
Charges entering 0.100 mm of axon is 
substituting the value of 
in above equation, we get charge enter a 0.100 mm length of the axon is
In this 2-dimensional graph, the x-component of each vector is the horizontal distance from the origin, while the y-component of each vector is the vertical distance from the origin. It can be seen that the c vector is 1 vertical unit away from the origin, which means that it has a y-component of 1.
let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
Complete question:
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The
particle is momentarily at rest at t is:
Select one:
a. 9.3s
b. 1.3s
C. 0.75s
d.5.3s
e. 7.3s
Answer:
b. 1.3 s
Explanation:
Given;
position of the particle, x(t)=1 6t- 3.0t³
when the particle is at rest, the velocity is zero.
velocity = dx/dt
dx /dt = 16 - 9t²
16 - 9t² = 0
9t² = 16
t² = 16 /9
t = √(16 / 9)
t = 4/3
t = 1.3 s
Therefore, the particle is momentarily at rest at t = 1.3 s
