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3 years ago

7

Are graded receptor potentials always depolarizing? Do graded receptor potentials always make it easier to induce action potenti

al?

1 answer:

Oksana_A [137]3 years ago

7 0

Answer: Yes,graded receptor potential always depolarize.

Yes,graded receptor potentials must occur to depolarize the neutrons to threshold before action potentials can occur.

Explanation:

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Why is there no effect on other branches in a paral- lel circuit when one branch of the circuit is opened or closed?

motikmotik

Answer:

Explanation:

As the circuit is parallel, then there is no effect of other branches as the potential difference across each arm is same.

6 0

3 years ago

An ice skater starts with a velocity of 2.25 m/s in a 50.0 degree direction. After 8.33s, she is moving 4.65 m/s in a 120 degree

Mnenie [13.5K]

The y-component of the acceleration is $0.22 m/s^2$

Explanation:

The y-component of the acceleration is given by

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

u = 2.25 m/s is the initial velocity, in a direction $\theta=50.0^{\circ}$

v = 4.65 m/s is the final velocity, in a direction $120^{\circ}$

t = 8.33 s is the time elapsed

The y-components of the initial and final velocity are:

$u_y = u sin \theta = (2.25)(sin 50^{\circ})=1.72 m/s\\v_y = v sin \theta = (4.65)(sin 50^{\circ})=3.56 m/s$

So the y-component of the acceleration is

$a_y = \frac{3.56-1.72}{8.33}=0.22 m/s^2$

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3 years ago

NEWTONS SECOND LAW LAB REPORT

hram777 [196]

Answer:

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

Explanation:

here this may help.

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3 years ago

Rocks from volcanoes are used by scientists to study the interior of the earth<br><br>TRUE OR FALSE

cupoosta [38]

I think it’s false bc A volcano is not in the interior of the earth

6 0

3 years ago

Read 2 more answers

You observe a hockey puck of mass 0.12 kg, traveling across the ice at speed 18.3 m/sec. The interaction of the puck and the ice

Galina-37 [17]

The stopping distance is 143.1 m

Explanation:

First of all, we have to find the acceleration of the hockey puck. This can be done by using Newton's second law of motion:

$\sum F =ma$

where

$\sum F = F_f = -0.14 N$ is the net force acting on the puck (the force of friction, negative because it acts in a direction opposite to the direction of motion)

m = 0.12 kg is the mass of the puck

a is the acceleration

Solving for a,

$a=\frac{\sum F}{m}=\frac{-0.14}{0.12}=-1.17 m/s^2$

The motion of the puck is a uniformly accelerated motion, therefore we can use the following suvat equation:

$v^2-u^2=2as$

where:

v = 0 is the final velocity (the puck comes to a stop)

u = 18.3 m/s is the initial velocity

$a=-1.17 m/s^2$ is the acceleration

s is the stopping distance

And solving for s, we find

$s=\frac{v^2-u^2}{2a}=\frac{0-(18.3)^2}{2(-1.17)}=143.1 m$

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3 years ago