Ask question

Ask question

All categories

3 years ago

8

Aplicación de las palancas en la Fisioterapia.

1 answer:

Anon25 [30]3 years ago

8 0

En términos simples, una articulación (donde dos o más huesos se unen) forma el eje (o fulcro), y los músculos que cruzan la articulación aplican la fuerza para mover un peso o resistencia. Las palancas suelen estar etiquetadas como de primera, segunda o tercera clase

You might be interested in

A ball falls from the top of a building. As it falls, its speed increases. Which type of energy is the ball gaining as it falls?

Elina [12.6K]

<h2>Hello!</h2>

The answer is: B. Kinetic energy

<h2>Why?</h2>

Since the ball is falling, speed increases because the gravity acceleration is acting. When speed increases, the kinetic energy increases too, so the ball is gaining kinetic energy.

The gravity acceleration is equal to $9.81\frac{m}{s^{2}}$ , it means that when falling, the ball will increase it's speed 9.81m every second.

We can calculate the kinetic energy by using the following formula:

$KE=\frac{1}{2}*m*v^{2}$

Where:

$m=mass\\v=velocity$

Have a nice day!

<h2 />

5 0

3 years ago

A 30° incline permanently sits on a 1.1 meter high table. Starting from rest a ball rolls off the incline with a velocity of 2m/

exis [7]

It would be a good game for you but if I get a pic I don’t want you can you come to my crib I just

3 0

3 years ago

1. A car is slowing down at a constant rate from 40 m/s to come to a stop. The acceleration is - 5 m/s2. How

Oxana [17]

Answer:

Explanation:

5 0

3 years ago

A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance,

Lisa [10]

Answer:

Explanation:

The two charges are q and Q - q. Let the distance between them is r

Use the formula for coulomb's law for the force between the two charges

$F = \frac{Kq_{1}q_{2}}{r^{2}}$

So, the force between the charges q and Q - q is given by

$F = \frac{K\left ( Q-q \right )q}}{r^{2}}$

For maxima and minima, differentiate the force with respect to q.

$\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )$

For maxima and minima, the value of dF/dq = 0

So, we get

q = Q /2

Now $\frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}$

the double derivate is negative, so the force is maxima when q = Q / 2 .

6 0

3 years ago

What is the net power that a person with surface area of 1.20 m2 radiates if his emissivity is 0.895, his skin temperature is 27

marissa [1.9K]

Explanation:

Below is an attachment containing the solution.

4 0

3 years ago