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Anni [7]
4 years ago
12

How do the two reflex arcs differ in complexity

Physics
1 answer:
DanielleElmas [232]4 years ago
3 0
<span>In a monosynaptic reflexes, it is like me speaking to you over a telephone. I am the only talker and you are the only listener. The phone line only has to connect to 2 people. If I am the one talking, I would be the sensory neuron, and if you are listening you would be the motor neuron. I can tell you what to do really fast because it is only 1 person I have to give instructions to. An example of this is the patellar reflex. The doc hits the patella and your leg kicks. This is becuse the sensory neuron gets hit and he quickly tells the motor neuron to kick. It happens really, really fast. Another example is touching a hot stove, what happens? Before you can even think to yourself that the stove is hot, your hand will pull away. Monosynaptic reflexes are made for very quick responses to stimuli that will not need to be stopped or thought about in order to react.



</span>
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A spring has a length of .200 m when a .300 kg hangs from it,and a length of .750 m when a 1.95 kg hangs from it.what is the for
musickatia [10]

1) 29.5 N/m

2) 0.100 m

Explanation:

1)

The force constant of the spring can be found by using the fact that the force on the spring is proportional to the extension of the spring (Hooke's Law). Therefore, we can write:

\Delta F= k \Delta x

where

\Delta F = F_2 - F_1 is the change in the force on the spring, where

F_1 = m_1 g = (0.300)(9.8)=2.94 N is the force applied when the hanging mass is

m_1 = 0.300 kg

F_2 = m_2 g = (1.95)(9.8)=19.1 N is the force applied when the hanging mass is

m_2 = 1.95 kg

\Delta x=x_2 -x_1 is the change in extension of the spring, where

x_1=0.200 m is the extension of the spring when the hanging mass is 0.300 kg

x_2=0.750 m is the extension of the spring when the hanging mass is 1.95 kg

Solving for k,

k=\frac{F_2-F_1}{x_2-x_1}=\frac{19.1-2.94}{0.750-0.200}=29.5 N/m

2)

When the first mass is hanging on the spring, we have

F_1 = k (x_1 - x_0)

where:

F_1 is the force applied on the spring (the weight of the hanging mass)

k is the spring constant

x_1 is the extension of the spring wrt its natural length

x_0 is the natural length of the spring (the unloaded length)

Here we have

F_1=2.94 N

k = 29.5 N/m

x_1=0.200 m

Solving for x_0, we find:

x_0 = x_1 - \frac{F_1}{k}=0.200 - \frac{2.94}{29.5}=0.100 m

5 0
3 years ago
What type of resource can be regenerated or replenished by biochemicl cycles
Valentin [98]

A: A resource that will always be there, can be replenished by the biogeochemical cycles. B: Can regenerate if they are alive or can be replenished by biochemical cycles if they are non living.

5 0
4 years ago
Which list contains only objects that orbit the sun in our solar system?
Andre45 [30]

The answer would be C.

3 0
4 years ago
An object oscillates with an angular frequency of 8.0 rad/s. At t = 0, the object is at x0 = 4 cm with an initial velocity v0 =
KatRina [158]

Answer:

\phi = 0.66 rad

A = 5.06 cm

Explanation:

We have here a simple harmonic motion, so the equation of the position in this motion is:  

x(t)=Acos(\omega t+\phi) (1)

A: Amplitude

ω: angular frequency

φ: phase constant

If we take the derivative of x with respect to t from (1), we can find the velocity equation of this motion:

v(t)=\frac{dx(t)}{dt}=-A\omega sin(\omega t+\phi) (2)

Let's evaluate (1) and (2) in t=0.

x(0)=Acos(\phi) (3)

v(0)=-A\omega sin(\phi) (4)

Dividing 4 by 3 we have:

\frac{v(0)}{x(0)}=-\omega tan(\phi)

\phi = tan^{-1}(\frac{-v(0)}{\omega x(0)})

\phi = 0.66 rad

Now, using (3) we can find the amplitude.

A = \frac{x(0)}{cos(\phi)} = 5.06 cm

I hope it helps!                            

6 0
4 years ago
Question #4
olga2289 [7]

Answer:

Distance 5 km, Displacement 3 km east

Explanation:

6 0
3 years ago
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