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2 years ago

Educación física como se puede mejorar la técnica deportiva

Physics

1 answer:

motikmotik2 years ago

8 0

Answer:

Varíe sus entrenamientos. Para muchos atletas, los ejercicios funcionales son el nombre del juego. ...

Realice un seguimiento y mida su rendimiento durante el entrenamiento. ...

Haga de la hidratación adecuada una prioridad. ...

Dedique suficiente tiempo a la recuperación. ...

Entrena tu cerebro. ...

Alimente su cuerpo de la manera correcta. ...

Considere agregar algunos suplementos a su dieta.

Explanation:

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HELP ME PLEASE!!!!!!!!!

Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

$F = \frac{kq_1q_3}{d_{13}^2}$

now from the above equation

$F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}$

$F = 0.288 N$

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

$F_{net} = 2Fcos\theta$

here we know that angle is

$\theta = 37 degree$

now we have

$F_{net} = 2\times 0.288 cos37$

$F_{net} = 0.46 N$

so net force on q3 is 0.46 N vertically upwards along +Y axis

6 0

2 years ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

2 years ago

Help me with this question Please

zysi [14]

Answer:

W has the lowest density and Y has the greatest density

Explanation:

Density of W = mass/volume = 11/24 = 0.45

Density of X = mass/volume = 11/12 = 0.91

Density of Y = m/v = 5.5/4 = 1.375

Density of Z = m/v = 5.5/11 = 0.5

From these we can find the answer......

Hope this answer is useful......

3 0

2 years ago

The surface area of a postage

ANTONII [103]

Answer:

608

Explanation:

Trust me

4 0

1 year ago

Read 2 more answers

The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr

s344n2d4d5 [400]

Answer:C.The tube should be held vertically, perpendicular to the ground.

Explanation: Power lines are mainly overhead power installation that transfer electric energy from one place to another. Electric power lines contains both Magnetic field and Electric field.

Potential difference is the change in the amount of energy carried by an electric circuit from one point to another. TO MAXIMIZE THE POTENTIAL DIFFERENCE BETWEEN ONE END OF THE TUBE AND ANOTHER? THE TUBE SHOULD BE HELD VERTICALLY,PERPENDICULAR TO THE GROUND.

7 0

2 years ago