3 similar things between HIIT,CrossFit y Bodyweight

Fofino [41]

Answer:

The retina is the light sensitive focusing screen. Inside, there's two types of fluid, one watery behind the lens, and further back, a thicker one that helps the eyeball hold its shape.

8 0

2 years ago

Jalen is sliding around on his kitchen floor in his

tresset_1 [31]

Answer:

his feet feel warmer because of friction. when friction takes place then an object will warm up because friction is a universal law that always takes place. if you rub your hands together, you also experience the heat of frictional I hope that helped.

5 0

3 years ago

If the net force acting on an object is zero, its inertia is also zero.<br> True or false

S_A_V [24]

Answer:

its false

Explanation:

5 0

3 years ago

Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is

suter [353]

Answer: $0.223 m/s^{2}$

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity $g$ of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

$F=G\frac{m_{E}m}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$m_{E}=5.98(10)^{24} kg$ is the mass of the Earth

$m$ are the mass of each communications satellite

$r=R_{E}+h$ is the distance between the center of the Earth and the satellite

$R_{E}=6.38(10)^{6} m$ is the radius of the Earth

$h=3.59(10)^{7} m$ is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to <u>Newton's 2nd law of motion:</u>

$F=mg$ (2)

Combining (1) and (2):

$G\frac{m_{E}m}{r^2}=mg$ (3)

Isolating $g$ :

$g=\frac{G M_{E}}{r^2}$ (4)

Remembering $r=R_{E}+h$ :

$g=\frac{G M_{E}}{(R_{E}+h)^2}$ (5)

$g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}$

Finally:

$g=0.223 m/s^{2}$

5 0

3 years ago

A 5.0 kg block hangs from the ceiling by a mass-less rope. A Second block with a mass of 10.0 kg is attached to the first block

gayaneshka [121]

The tension in the first and second rope are; 147 Newton and 98 Newton respectively.

Given the data in the question

Mass of first block; $m_1 = 5.0kg$

Mass of second block, $m_2 =10kg$

Tension on first rope; $T_1 =\ ?$

Tension on second rope; $T_2 =\ ?$

To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

$F = m\ *\ a$

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity $g = 9.8m/s^2$ )