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3 years ago

Identify the four main divisions of the peripheral nervous system and explain each one's function.

Physics

2 answers:

Nataly [62]3 years ago

8 0

Answer:

That's correct ^-^

guajiro [1.7K]3 years ago

3 0

Answer:

The peripheral nervous system is divided into the somatic and autonomic nervous systems, and the autonomic nervous system is divided into the sympathetic and parasympathetic nervous systems. The somatic nervous system controls voluntary skeletal muscles. The autonomic nervous system controls the glands and the muscles of the internal organs. The sympathetic nervous system arouses the body and expends energy. It is responsible for our fight and flight response. The parasympathetic nervous system calms the body and conserves energy. It is responsible for our rest and digest response.

Explanation:

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What does the length of a vector arrow represent?

nalin [4]

The length of a vector arrow represents an magnitude

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1 year ago

A system gains 757 kJ757 kJ of heat, resulting in a change in internal energy of the system equal to +176 kJ.+176 kJ. How much w

Alecsey [184]

Answer:

581 kJ, work was done by the system

Explanation:

According to the first law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the system

Q is the heat absorbed by the system (positive if absorbed, negative if released)

W is the work done by the system (positive if done by the system, negative if done by the surrounding)

In this problem,

$Q=+757 kJ$

$\Delta U = +176 kJ$

Therefore the work done by the system is

$W=Q-\Delta U=757 kJ-176 kJ=+581 kJ$

And the positive sign means the work is done BY the system.

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2 years ago

A wire that is 0.65 m long and carrying a current of 8.2 A is at right angles to a uniform magnetic field. The force on the wire

Luda [366]

Answer:

0.075 T

Explanation:

When a current-carrying wire is immersed in a region with magnetic field, the wire experiences a force, given by

$F=ILB sin \theta$

where

I is the current in the wire

L is the length of the wire

B is the strength of the magnetic field

$\theta$ is the angle between the direction of I and B

In this problem we have:

L = 0.65 m is the length of the wire

I = 8.2 A is the current in the wire

F = 0.40 N is the force experienced by the wire

$\theta=90^{\circ}$ since the current is at right angle with the magnetic field

Solving the formula for B, we find the strength of the magnetic field:

$B=\frac{F}{IL sin \theta}=\frac{0.40}{(8.2)(0.65)(sin 90^{\circ})}=0.075 T$

3 0

3 years ago

A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?

garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

$U_i +K_i = U_f + K_f$

where:

$U_i$ is the initial potential energy, at the bottom

$K_i$ is the initial kinetic energy, at the bottom

$U_f$ is the final potential energy, at the top

$K_f$ is the final kinetic energy, at the top

We can rewrite the equation as:

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m is the mass of the hopper

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 0$ is the initial height

u is the launch speed of the hopper

$h_f = 1.20 m$ is the maximum altitude reached by the hopper

v = 0 is the final speed (which is zero when the hopper reaches the maximum height)

Solving the equation for u, we find the launch speed of the hopper:

$u=\sqrt{2gh_g}=\sqrt{2(9.8)(1.20)}=4.8 m/s$

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0

2 years ago

An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.

Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

$The \ weight, W =G\dfrac{M \times m}{R^{2}} = 800 \ N$

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

$W_1 =G\dfrac{\dfrac{M}{2} \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2} \times m \times 4}{ R ^{2}} = 2 \times G \times \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N$

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0

2 years ago