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2 years ago

Aplicación de las palancas en la Fisioterapia.

1 answer:

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

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A flashlight bulb with a 6.00 resistor uses 18.0W of power. What is the current through the bulb

devlian [24]

We need voltage first

P=V²/R
V²=PR
V²=18(6)
V²=108
V=√108
V=10.4V

Apply oHMS law

V/I=R
I=V/R
I=10.4/6
I=1.73A

6 0

2 years ago

Read 2 more answers

A balloon contains 0.075 m^3 of

inessss [21]

Answer:

600 KPa.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V1) = 0.075 m³

Final volume (V2) = 0.45 m³

Final pressure (P2) = 100 KPa

Initial pressure (P1) =?

Temperature = constant

The initial pressure can be obtained by using the Boyle's law equation as shown below:

P1V1 = P2V2

P1 × 0.075 = 100 × 0.45

P1 × 0.075 = 45

Divide both side by 0.075

P1 = 45 / 0.075

P1 = 600 KPa.

Thus, the initial pressure in the balloon is 600 KPa.

7 0

2 years ago

What do we call the energy that is transferred to

mojhsa [17]

Answer:

Latent heat

Latent heat is the heat needed to change a state without a change in temperature.

Explanation:

Hope this helps :D

5 0

2 years ago

1) What would the average acceleration be for a car at a stoplight that speeds up to 20 m/s in 10 seconds (in m/s^2)

Alexxx [7]

1.)
Velocity is in m/s, and acceleration is in m/s^2 like you said. Because of this, we can calculate this by dividing the speed by the time it took to get to that speed.
(20 meters/second) / 10 seconds = 2 meters/ second^2

2.)
Same thing with the first one.
(100 meters/second) / 4 seconds = 25 meters / seconds^2

7 0

2 years ago

Read 2 more answers

Find the ratio of the new/old periods of a pendulum if the pendulum were transported from earth to the moon, where the accelerat

vichka [17]

The period of a pendulum is given by
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length and g is the gravitational acceleration.

We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:
$\frac{T_m}{T_e} = \frac{2 \pi \sqrt{ \frac{L}{g_m} } }{2 \pi \sqrt{ \frac{L}{g_e} } }= \sqrt{ \frac{g_e}{g_m} }$
where the labels m and e refer to "Moon" and "Earth".

Since the gravitational acceleration on Earth is $g_e = 9.81 m/s^2$ while on the Moon is $g_m=1.63 m/s^2$ , the ratio between the period on the Moon and on Earth is
$\frac{T_m}{T_e}= \sqrt{ \frac{g_e}{g_m} }= \sqrt{ \frac{9.81 m/s^2}{1.63 m/s^2} }=2.45$

3 0

2 years ago