PHGY 214 Lecture Notes - Lecture 2: Amacrine Cell, Glion

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12 May 2016

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Direction of motion what is the first neuron in the reti- na that has the ability to generate action poten- tials. Amacrine cells lie right between bipolar cells and ganglion cells. Information from the bipo- lar cell, rec- which is not di- tion- ally sensi- becomes tive, directionally sensitive in the amacrine cell. Amacrine cells have the ability to go in many directions (multi-directional selectivity) There are many types of off-centre and on-centre cells. However, the location of the axon terminals differ they also differ in functions as well the time it takes to respond to light. Where do these types of bipolar cells terminate on the dendritic trees. Bc2 has a delay in the response and bc3a activates right away. When bc2 and bc3a respond at the same time and you get a much bigger depolarization. If you reverse the direction, they don t happen at the same time, so there is not as large of a depolarization.

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This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page - find all choices before answering. Two vectors A and B lie in the xy plane and are given by A = Ax i + Ay j B = Bx i + By j, where Ax = 3.58 m, Ay = 2.85 m, Bx = 7.09 m, By = -6.36 m. Find the magnitude of R = A + B. Answer in units of m Find the angle theta (between -180 degree and + 180 degree with counterclockwise positive) that the vector R makes from the positive x axis. Answer in units of degree When the Sun is directly overhead, a hawk dives toward the ground at a speed of 7.23 m/s. If the direction of his motion is at an angle of 39.8 below the horizontal, calculate the speed of his shadow along the ground. Answer in units of m/s Vector A has a magnitude of 10 and points in the positive x-direction, Vector B has a magnitude of 17 and makes an angle of 32 degree with the positive x-axis. What is the magnitude of A - B? An airplane starting from airport A (lies 342 km east, then 531 km at 62.2 degree west of north, and then 778 km north to arrive finally at airport B. The next day, another plane flies directly from A to B in a straight line. In what direction should the pilot trawl in this direct flight? Use counterclockwise its the positive angular direction from due east, between the limits of - 180 degree ami 180 degree . Assume there is no wind during these flights. Answer in units of degree How far will the pilot travel in this direct flight? Answer in units of km A projectile is fired straight upward at 126 m/s. How fast is it moving at the instant it reaches the top of its trajectory? Answer in units of m/s How fast is it moving at the instant it reaches the top of its trajectory if the projectile is fired upward at 12 degree from the horizontal? Answer in units of m/s At one instant the velocity of a particle is and a very short time later its velocity has changed to . Qualitatively, what was the direction of the particle's acceleration during the interval between the first and second instance?

which of the vectors below best represents the direction of the impulse vector j⃗ ?

To learn about the impulse-momentum theorem and its applications in some common cases.

Using the concept of momentum, Newton's second law can be rewritten as

?F? =dp? dt, (1)

where ?F? is the net force F? net acting on the object, and dp? dt is the rate at which the object's momentum is changing.

If the object is observed during an interval of time between times t1 and t2, then integration of both sides of equation (1) gives

?t2t1?F? dt=?t2t1dp? dtdt. (2)

The right side of equation (2) is simply the change in the object's momentum p2??p1?. The left side is called the impulse of the net force and is denoted by J? . Then equation (2) can be rewritten as

J? =p2??p1?.

This equation is known as the impulse-momentum theorem. It states that the change in an object's momentum is equal to the impulse of the net force acting on the object. In the case of a constant net force F? net acting along the direction of motion, the impulse-momentum theorem can be written as

F(t2?t1)=mv2?mv1. (3)

Here F, v1, and v2 are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in question is two-dimensional, it is often useful to apply equation (3) to the x and y components of motion separately.

( The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass m moving along the x-axis. The net force F is acting on the particle along the x-axis. F is a constant force.)

A) The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0.

B) The particle starts from rest at t=0. What is the magnitude v of the velocity of the particle at time t? Assume that t>0.

C) The particle has the momentum of magnitude p1 at a certain instant. What is p2, the magnitude of its momentum ?t seconds later?

D) The particle has the momentum of magnitude p1 at a certain instant. What is v2, the magnitude of its velocity ?t seconds later?

E) Find the magnitude of the impulse J delivered to the particle.

F) Which of the vectors below best represents the direction of the impulse vector J??
(Figure 1)

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puceworm982

In Example 33.11, the water-glass interface is horizontal. If instead this interface were tilted 15.0° above the horizontal, with the right side higher than the left side, what would be the angle from the vertical of the ray in the glass? (The ray in the water still makes an angle of 60.0° with the vertical).

EXAMPLE 33.11: Reflection and refraction

In Fig 33.11, material a is water and material b is a glass with an index of refraction 1.52. If the incident ray makes an angle of 60.0° with the normal, find the directions of the reflected and refracted rays.

SOLUTION

IDENTIFY: This is a problem in geometric optics. We are given the incident angle and the index of refraction of each material, and we need to find the reflected and refracted angles.

SET-UP: Figure 33.11 shows the rays and angles for this situation. The target variables are the reflected angle θ_r and the refracted angle θ_b. Since n_b is greater than n_a, the refracted angle must be smaller than the incident angle θ_a; this is shown in the figure.

EXECUTE: According to Eq 33.2, the angle the reflected ray makes with the normal is the same as that of the incident ray so θ_r= θ_a=60.0°.

To find the direction of the refracted ray, we use Snell’s law, Eq 33.4, with n_a=1.33, n_b=1.42, and θ_a=60.0°. We find:

EVALUATE: The second material has a larger refractive index than the first, just like the situation shown in Fig 33.8a. Hence, the refracted ray is bent toward the normal as the wave slows down upon entering the second material and θ_b< θ_a.

Figure 33.11. Reflection and refraction of light passing from water to glass

Figure 33.8. Reflection and refraction in three cases. (a) Material b has a larger index of refraction than material a. (b) (a) Material b has a smaller index of refraction than material a. (c) The incident light ray is normal to the interface between the materials.

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PHGY 214 Lecture Notes - Lecture 2: Amacrine Cell, Glion

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