All categories

3 years ago

What are the three jobs of the nervous system

1 answer:

fomenos3 years ago

5 0

The nervous system has three general functions: a sensory function, an interpretative function and a motor function. 1. Sensory nerves gather information from inside the body and the outside environment. The nerves then carry the information to the central nervous system (CNS). 2. Sensory information brought to the CNS is processed and interpreted. 3. Motor nerve cells convey information from the CNS to the muscles and glands of the body. The nervous system is responsible for coordinating all of the body's activities.

You might be interested in

An LC circuit consists of a 3.4-µF capacitor and a coil with a self-inductance 0.080 H and no appreciable resistance. At t = 0 t

alexira [117]

Answer

given,

capacitance = C = 3.4-µF

inductance = L = 0.08 H

frequency is expressed as

$f = \dfrac{1}{2\pi\sqrt{LC}}$

time period

$T = \dfrac{1}{f}=2\pi\sqrt{LC}$

after time T/4 current reach maximum

$t = \dfrac{T}{4}$

$t = \dfrac{2\pi\sqrt{LC}}{4}$

$t = \dfrac{2\pi\sqrt{0.08 \times 3.4 \times 10^{-6}}}{4}$

t = 8.2 x 10⁻⁴ s

t = 0.82 ms

b) using law of conservation

$\dfrac{1}{2}CV^2=\dfrac{1}{2}LI^2$

$I^2 = \dfrac{CV^2}{L}$

$I^2 = \dfrac{C}{L}\dfrac{Q^2}{C^2}$

$I =\sqrt{\dfrac{Q^2}{CL}}$

$I =\sqrt{\dfrac{(5.4 \times 10^{-6})^2}{0.08 \times 3.4 \times 10^{-6}}}$

I = 0.010 A

I = 10 mA

4 0

3 years ago

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

2 years ago

Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missil

tatyana61 [14]

Answer:

The time taken by missile's clock is $4.6\times 10^{6} s$

Solution:

As per the question:

Speed of the missile, $v_{m = 6.5\times 10^{3}} m/s$

Now,

If 'T' be the time of the frame at rest then the dilated time as per the question is given as:

T' = T + 1

Now, using the time dilation eqn:

$T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}$ (1)

Using binomial theorem in the above eqn:

We know that:

$(1 + x)^{a} = 1 + ax$

Thus eqn (1) becomes:

$1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}$

$T = \frac{2c^{2}}{v_{m}^{2}}$

Now, putting appropriate values in the above eqn:

$T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}$

$T = 4.6\times 10^{6} s$

4 0

3 years ago

At what condition does a body becomes weightless at the equator?

ArbitrLikvidat [17]

Answer:

The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface

Explanation:

4 0

2 years ago

To properly operate the laboratory thermometer Group of answer choices it needs to be shaken but make sure you have enough room

ruslelena [56]

Answer:

it needs to be shaken but make sure you have enough room to shake it safely

Explanation:

To properly operate the laboratory thermometer it needs to be shaken but make sure you have enough room to shake it safely. This done because there is a small bend in the mercury channel of a clinical thermometer that uses mercury. You must shake the thermometer to get the mercury from a previous reading from the thermometer back into the bulb for taking new reading. The bend prevents flow back into the tube so that one can comfortably take reading.

5 0

3 years ago