Ask question

Ask question

All categories

3 years ago

14

Que capacidade física consiste em deslocar o corpo no espaço o mais rápido possível, mudando o centro de gravidade de posição, s

em perder o equilíbrio e a coordenação dos movimentos?

1 answer:

grin007 [14]3 years ago

6 0

Answer:ur mom

Explanation:because

You might be interested in

What is the minimum amount of data points an experiment should gather?

nydimaria [60]

The answer is "Three".

At the point when individuals take studies for factual purposes or when the Statistics Agency comes around and gets everyone's data, data from one individual is one information point for them. Toward the finish of its exploration or overview, the organization will have accumulated numerous bits of data from numerous individuals. One piece of information equals with one data point.

7 0

3 years ago

Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6

Nana76 [90]

Answer:

The answer to the question is as follows

The acceleration due to gravity for low for orbit is 9.231 m/s²

Explanation:

The gravitational force is given as

$F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }$

Where $F_{G}$ = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹ $\frac{Nm^{2} }{kg^{2} }$

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by $F_{G}$ and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and $F_{G}$ = $\frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} }$ = 9.231 N

Therefore the acceleration due to gravity = $F_{G}$ /mass

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²

4 0

3 years ago

Read 2 more answers

Which positions experience low tide

MrMuchimi

Answer:

lentic zones because they do not move they are stagnant water examples are ponds, swap, dams.

8 0

3 years ago

A skateboarder shoots off a ramp with a velocity of 7.1 m/s, directed at an angle of 61° above the horizontal. The end of the ra

dsp73

Answer:

Highest point reached = 3.37 m

Explanation:

Initial velocity, = 7.1 m/s

Initial vertical velocity = 7.1 sin 61 = 6.21 m/s

Consider the vertical motion of skateboarder,

We have equation of motion, v² = u² + 2as

Initial velocity, u = 6.21 m/s

Acceleration, a = -9.81 m/s²

Final velocity, v = 0 m/s

Substituting

v² = u² + 2as

0² = 6.21² + 2 x -9.81 x s

s = 1.97 m

So from ramp the position it goes up by 1.97 m

Highest point reached = 1.97 + 1.4 = 3.37 m

6 0

3 years ago

A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o

likoan [24]

Answer:

$U_{b}=+7.3*10^{-8}J$

Explanation:

Given data

Electric potential at point a is Ua=5.4×10⁻⁸J

q₂ moves to point b where a negative work done on it $W_{a-b}=-1.9*10^{-8}J$

Required

Electric potential energy Ub

Solution

When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:

$W_{a-b}=U_{a}-U_{b}\\U_{b}=U_{a}-W_{a-b}$

Now substitute the given values

So

$U_{b}=5.4*10^{-8}J-(-1.9*10^{-8}J)\\U_{b}=+7.3*10^{-8}J$

3 0

3 years ago