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2 years ago

Which structure reduces friction between joints?

Physics

1 answer:

Roman55 [17]2 years ago

3 0

Answer:

The cartilage is the structure that reduces friction between joints.

Explanation:

Cartilage is the tissue between the bone joints. They are composed of several membranes and cellular groups. One of them called synovial membrane which provides lubricant to the edges of tissue that are touched by the bones. Reducing the friction and absorbing it so the body can move without obstacles.

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Which law states that absolute zero cannot be reached?

padilas [110]

The third law of thermodynamics,the principle of temperature.

This law states that the entropy at 0 is always equel to 0.
This means that it is impossible to cool down a perfect 0 or absolute 0(-273.15 C)

5 0

3 years ago

Read 2 more answers

011 10.0 points

Ulleksa [173]

Answer:

2.47 m

Explanation:

Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.

The horizontal velocity of the ball is constant:

$v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s$

and the time taken to cover the horizontal distance d is

$t=\frac{d}{v_x}=\frac{52}{20.3}=2.56 s$

So this is the time the ball takes to reach the horizontal position of the crossbar.

The vertical position of the ball at time t is given by

$y=u_y t - \frac{1}{2}gt^2$

where

$u_y = v sin \theta =(25)(sin 35.9^{\circ})=14.7 m/s$ is the initial vertical velocity

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

And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:

$y=(14.7)(2.56) - \frac{1}{2}(9.8)(2.56)^2=5.52 m$

The height of the crossbar is h = 3.05 m, so the ball passes

$h' = 5.52- 3.05 = 2.47 m$

above the crossbar.

8 0

3 years ago

Corresponde al conjunto de puntos por donde pasa un cuerpo al moverse

Anna35 [415]

Answer:

Means?

Explanation:

6 0

3 years ago

A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical

iris [78.8K]

Answer:

17.71N/m

Explanation:

The period of the spring is expressed according to the expression;

$T = 2 \pi \sqrt{\frac{m}{k} } \\$

m is the mass of the object

k is the force constant

Given

m = 5.50kg

T = 3.50s

Substitute into the formula;

$T = 2 \pi \sqrt{\frac{m}{k} } \\3.5 = 2 (3.14) \sqrt{\frac{5.5}{k} } \\3.5 = 6.28 \sqrt{\frac{5.5}{k} } \\\frac{3.5}{6.28} = \sqrt{\frac{5.5}{k} } \\0.557 = \sqrt{\frac{5.5}{k} } \\square \ both \ sides\\0.557^2 = (\sqrt{\frac{5.5}{k} })^2 \\0.3106 = \frac{5,5}{k}\\k = \frac{5.5}{0.3106}\\k = 17.71N/m$

Hence the force constant of the spring is 17.71N/m

4 0

3 years ago

When two point charges are a distance d apart, the magnitude of the electrostatic force between them is F. If the distance betwe

EastWind [94]

Answer:

The magnitude of the electrostatic force decreases by a factor 9

Explanation:

The electrostatic force between two charges is given by:

$F=k\frac{q_1 q_2}{d^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

d is the distance between the two charges

We see that the magnitude of the force is inversely proportional to the square of the distance. If the distance is increased to 3d: d' = 3d, the new electrostatic force would be:

$F'=k\frac{q_1 q_2}{(d')^2}=k\frac{q_1 q_2}{(3d)^2}=\frac{1}{9} k\frac{q_1 q_2}{d^2}=\frac{F}{9}$

So, the electrostatic force decreases by a factor 9.

3 0

3 years ago

Read 2 more answers