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Tpy6a [65]
3 years ago
9

1. Describe the components of the reflex arc

Physics
1 answer:
m_a_m_a [10]3 years ago
4 0
The simplest arrangement of a reflex arc consists of the receptor, an interneuron (or adjustor), and an effector; together, these units form a functional group. Sensory cells carry input from the receptor (afferent impulses) to a central interneuron, which makes contact with a motor neuron.
You might be interested in
A simple harmonic oscillator completes 1550 cycles in 30 min. (a) Calculate the period. s (b) Calculate the frequency of the mot
nordsb [41]

Answer:

(a) 1.16 s

(b)0.861 Hz

Explanation:

(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.

From the question,

If 1550 cycles is completed in (30×60) seconds,

1 cycle is completed in x seconds

x = 30×60/1550

x = 1.16 s

Hence the period is 1.16 seconds.

(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).

Mathematically, Frequency is given as

F = 1/T ........................... Equation 1

Where F = frequency, T = period.

Given: T = 1.16 s.

Substitute into equation 1

F = 1/1.16

F = 0.862 Hz

Hence thee frequency = 0.862 Hz

6 0
4 years ago
N in-ground swimming pool has the dimensions shown in the drawing. It is filled with water to a uniform depth of 3.00 m. The den
Doss [256]

Answer:

The pressure is  P = 1.31*10^{5} \ Pa

Explanation:

From the question we are told that

    The depth of the swimming pool is  d =  3.00 \  m

     The density of water is  \rho = 1.00*10^{3} \  kg /m^3

Generally the  total pressure exerted on the bottom of the swimming pool is mathematically represented as

          P = P_o + \rho * g * h

Here P_o is the atmospheric pressure with value

        P_o  =  101325 \  Pa

So

        P = 101325 + [1000 * 9.8 * 3]

=>     P = 130725 \ Pa

=>    P = 1.31*10^{5} \ Pa

     

7 0
3 years ago
During exercise, the amount of ____________ carried in the blood to the muscles increases.
ipn [44]
<span>The correct answer is: Oxygen

Explanation:
In order to function properly (movement etc.) during exercise, muscles require oxygen. During exercise, the depth as well as the rate of breathing increase, which in turn increases the amount of oxygen inhaled. In order to expand and contract lungs and for other bodily movements, muscles require oxygen, and for that, more oxygen is carried in the blood to muscles. Hence, the correct answer is Oxygen.</span>
8 0
3 years ago
Read 2 more answers
Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of
Pachacha [2.7K]

Answer:

<em>The frequency changes by a factor of  0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }

where f is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

f_{n} = \frac{1}{2\pi } \sqrt{\frac{k}{14m} }

simplifying, we have

f_{n} = \frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }

f_{n} = \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }

if we divide this final frequency by the original frequency, we'll have

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  ÷  \frac{1}{2\pi } \sqrt{\frac{k}{m} }

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  x  2\pi \sqrt{\frac{m}{k} }

==> 1/3.742 = <em>0.27</em>

7 0
3 years ago
A body with an initial velocity of 10m/s has an acceleration of 8m/s^2. Determine graphically the velocity after 5 seconds, &amp
Pie

Answer:

From the graph, at t = 5 seconds, the velocity = 50 m/s as shown also in the above table

Please find attached the graph

Explanation:

The initial velocity of the body =  m/s

The acceleration of the body = 8 m/s²

The velocity after 5 seconds can be determined graphically and by calculation as follows;

Graphically, we have the data points which can be found by the straight line relation  v = u + a×t,

Where ,

a = The slope = 8 m/s²

u = 10 m/s = The y-intercept

Which gives;

v = 10 + 8 × t

The following data can be calculated for various time t;

Time, t     Velocity , v

0,             10

1,              18

2,             26

3,             34

4,             42

5,             50

6,             58

From the graph, at t = 5 seconds, the velocity = 50 m/s as shown also in the above table

Please find attached the graph

By calculation, we have;

v = u + a×t

Where;

v = The final velocity

u = The initial velocity = 10 m/s

a = The acceleration = 8 m/s²

t = The time = 5 seconds

v = 10 + 5× 8 = 50 m/s.

6 0
3 years ago
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