The options of the given are:
A. A large diameter myelinated fiber
B. A small diameter myelinated fiber
C. A large unmyelinated fiber
D. A small unmyelinated fiber
E. A small fiber with multiple Schwann cells
Answer: Option A, A large diameter myelinated fiber.
Explanation:
The conduction of the nerve impulse would be greatest in the myelinated fiber because the main function of the myelin sheath is to increase the speed of the impulse at which the electrical signals propagate.
In case of the unmyelinated sheath the nerve impulse travels slowly as the conduction waves but in case of the large diameter myelinated sheath the signals travel via saltatory conduction( hop)
In this type of propagation the signals are transferred from the node of Ranvier in one neuron to next node which increases the overall velocity of the action potentials.
Answer:
A
Explanation:
When you pull the string the weight will experience a downward force and come closer to the vertical tube. And the angle between weight and hand through string is decreased so the angular velocity decreases
-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.
-- We know that the y-component of velocity is the derivative of the
y-component of position.
-- We're given the y-component of position as a function of time.
So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.
Now, the position function may look big and ugly in the picture. But with the
exception of 't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation. The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.
From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶
First derivative . . . y' (t) = (a₀ - g) t - 6 (a₀ / 30t₀⁴ ) t⁵ = (a₀ - g) t - (a₀ / 5t₀⁴ ) t⁵
There's your velocity . . . /\ .
Second derivative . . . y'' (t) = (a₀ - g) - 5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) - (a₀ /t₀⁴ ) t⁴
and there's your acceleration . . . /\ .
That's the one you're supposed to graph.
a₀ is the acceleration due to the model rocket engine thrust
combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀ is how long the model rocket engine burns
Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.
The big name in model rocketry is Estes. Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.
Answer:
that is like trying to learn your abc's at the age of 2 months old..........
Explanation:
:) just sayin
Answer:
How much power is required to pull a sled if you use 60J of work in 5 seconds? ... Misfortune occurs and Renatta and her friends find themselves getting a workout. They apply a cumulative force of 1080 N to push their car 218 m to the ... Calculate the amount of work done when moving a 567N crate a distance of 20 meters.
Explanation:
Misfortune occurs and Renatta and her friends find themselves getting a ... They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel ... Write down what they give you. ... Determine the work done by Lamar in deadlifting 300 kg to a height of 0.90 m ... Work = Force x Distance = Joules (force = weight).