un cuerpo suspendido en el aire es capaz de:
a) liberar energía al caer
b) puede ejecutar un trabajo
c) posee energía antes de caer
d) todas son correctas

The definition of the center of mass allows to find the result for the position of the mass center of more than the H₂O molecule is;

         [tex]x_{cm} = 0 \ and \ y_{cm} = 6.9 10^{-3 } nm[/tex]  

the concept of center of mass of a system is the point where external forces are applied, it is given by the expression

             [tex]\frac{x}{y} =\frac{1}{M_{total}} \sum m_i r_i[/tex]  

Where M is the total mass of the systemr_i and m_i sums the position and masses of the element i of the system

In the attachment we have a diagram of the system where the axis and coordinates of the molecules are shown, in this case it is indicated that the origin is in the oxygen atom, so its distance is zero.

           [tex]r_{cm} = \frac{1}{2 m + M} \ (2 m r )[/tex] )

They indicate the mass of the hydrogen atom m = 1.008 amu, the bond length r = 0.103 nm and there is an angle 106º between the two hydrogens, therefore the angle from the vertical is:

           θ = 106/2 = 53º

Let's find the position of the center of mass for each axis.

x-axis

           [tex]x_{cm} = \frac{1}{2m+ M} \ ( m x_1 - m x_2)[/tex]  

y-axis

          [tex]y_{cm} = \frac{1}{2m + M} \ ( m y_1 + m_2)[/tex]

Let's use trigonometry to find the components of the bond length.

         cos θ = [tex]\frac{y}{L}[/tex]  

         sin θ = [tex]\frac{x}{L}[/tex]  

         y = L cos θ

         x = L sin θ  

We substitute.

          [tex]x_{cm} = \frac{1}{2m+M} \ (mL (sin 53 + sin (-53)) \\y_{cm} = \frac{1}{2m + M} \ ( mL cos 53 + mL sin 53)[/tex]

we use.  

          sin θ = - sin -θ

          cos θ = cos -θ

Let's calculate.

        [tex]x_{cm} = 0 \\y_{cm} = \frac{1}{2 \ 1.008 + 15.9999} \ ( 2 \ 1.008 \ 0.103 cos 53)[/tex]

We see that the center of mass is on the x axis and at a distance from the y-axis of 6.9 10-3 nm

In conclusion using the definition of the center of mass we can find the result for the position of the center of mass of the H₂O molecule is;

          [tex]x_{cm}=0[/tex]  and [tex]y_{cm}[/tex]cm = 6.9 10⁻³ nm

Learn more here: brainly.com/question/8662931

Answer:

10

Explanation:

if you put 10 your right

Hi there!

A)

The angle that will produce the same range is the compliment of 35°.

Thus, kicking the ball at 55° will result in the same range.

We can prove this by using the derived range equation:

[tex]R = \frac{v^2sin2\theta}{g}[/tex]

An angle of 35° yields:

[tex]R = \frac{v^2sin(2*35)}{g} = .939R[/tex]

An angle of 55° yields:

[tex]R = \frac{v^2sin(2*55)}{g} = .939R[/tex]

Both are the same, thus indicating that 55° produces the same range.

B)

The angle of 55° will produce the higher ball because the VERTICAL component of the ball's velocity is greater compared to kicking the ball at 35° degree.

sin(55) > sin(35)

C)

The angle of 55° will result in the ball being in the air the longest because when a ball is in the air (assuming no air resistance), the ball experiences an acceleration due to gravity of -9.8 m/s², causing the vertical velocity to decrease until it eventually reaches 0 m/s at the top of its path. A greater initial vertical velocity means that it will take longer for the ball to fall.

We can prove this using:

vf = vi + at

0 = vy - 9.8t

vy/9.8 = t

Greater vy (vertical component of velocity) ⇒ greater time taken.

D)

The angle that would result in the furthest range is 45°.

We can prove this using calculus. Recall the above range equation:

[tex]R = \frac{v^2sin2\theta}{g}[/tex]

We can take the derivative and use the first-derivative test to find its critical point:

[tex]\frac{dR}{d\theta} = \frac{v^22cos2\theta}{g} = 0[/tex]

Evaluate:

[tex]v^22cos2\theta = 0 \\\\cos2\theta = 0 \\\\2\theta = 90^o\\\\\boxed{\theta = 45^o}[/tex]

A. The angle that will produce the same range as a 35° kick is 53°.

B. The angle that produces the highest ball is 45°.

C. The angle that produces the ball that is in the air the longest is 45°.

D. There is no single angle that will produce the furthest range for all initial velocities and accelerations due to gravity.

A. The range of a projectile is the horizontal distance it travels before it hits the ground. The range of a projectile is determined by the initial velocity, the angle of projection, and the acceleration due to gravity.

For a given initial velocity, the range of a projectile is maximized when the angle of projection is 45°. However, if the ground is not level, the range of a projectile can be maximized at other angles.

In the case of a soccer ball kicked at an angle of 35°, the range will be maximized at an angle of 53°. This is because the range of a projectile is maximized when the vertical component of the initial velocity is equal to the horizontal component of the initial velocity.

The angle of 53° is the angle that produces a vertical component of the initial velocity that is equal to the horizontal component of the initial velocity when the ball is kicked at an angle of 35°.

Therefore, the angle that will produce the same range as a 35° kick is 53°.

B. The angle that produces the highest ball is 45°.

C. The angle that produces the ball that is in the air the longest is 45°.

D. The angle that produces the furthest range depends on the initial velocity and the acceleration due to gravity. There is no single angle that will produce the furthest range for all initial velocities and accelerations due to gravity.

To know more about angle here

https://brainly.com/question/30147425

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this is my attachment answer hope it's helpful to you

Answer:

101.937 kg

Explanation:

The force needed to get the box moving must just cancel the static friction force:

F = µsmg  =  0.35•m•9.81 = 350 --->   m = 350 / (0.35•9.81) = 101.937 kg

Again, with units shown, and using 1 N = 1 kg•m/s2:

0.35•m•9.81(m/s2) = 350 N

solving for m:

m = 350 N / (0.35•9.81 m/s2)  =  (350 kg•m/s2) / (3.434 m/s2 ) = 101.937 kg

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(Hope this helps can I pls have brainlist (crown)☺️