Answer:
When the different ions cross the neuron membrane due to this an action potential generated
Explanation:
When the different ions cross the neuron membrane due to this an action potential generated.
The steps for action potential
1. Depolarization .When positively charge sodium ions enters in to the nerve cell.This is also known as upswing.
2. Repolarization
3. Refractory Phase
So the answer is "When the different ions cross the neuron membrane".
Answer:
A. The force exerted by the sprinter must be 9.6 × 10² N.
B. The force that propels the sprinter is exerted by the blocks.
Explanation:
Hi there!
Let´s begin with part B:
The sprinter exerts a force on the blocks and, as a reaction, the blocks exert a force on the sprinter that is of equal magnitude but opposite direction (Newton´s third law). This reaction of the blocks causes the acceleration of the sprinter.
Part A
The force exerted by the blocks can be calculated using Newton´s second law:
F = m · a
Where:
F = exerted force.
m = mass of the object being accelerated.
a = acceleration of the object after applying the force on the object.
F = m · a
F = 64 kg · 15 m/s²
F = 9.6 × 10² N
Period is defined as the time
required to complete one cycle which is also related to frequency. Period is
presented by an equation P = 1/f where p is the period, and f is the frequency.
The unit is in second. Frequency is defined as the number of circular
revolutions in a given time interval and it is in Hertz (Hz) = 1s^-1. Velocity
s defined as the arc traveled at a given time with the equation v=l/t where v
is the velocity, l is the length and t is time. The unit is in m/s.
The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.
The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)
−gsinθ=lθ¨−gainθ=lθ¨
For small angles, θ≪1,θ≪1, and hence sinθ≈θsinθ≈θ. Hence,
θ¨=−glθθ¨=−glθ
This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.
For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause