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3 years ago

The aorta pumps blood out of the heart to the rest of the body. Please select the best answer from the choices provided. T or f

2 answers:

3 0

The left ventricle relaxes and fills up with blood before squeezing and pumping the oxygen-rich blood through the aortic valve into the aorta — the main artery that carries blood to your body.<u>TRUE</u>

lisov135 [29]3 years ago

3 0

Answer:

True

Explanation:

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3 years ago

Are tsunamis earthquakes, or are they CAUSED by earthquakes?

neonofarm [45]

Answer:son causados por terremotos, ya que la fuerza de la tierra hace movimiento en las placas tectonicas debajo del mar creando asi los tsunamis.

Explanation:

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2 years ago

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You are at the controls of a particle accelerator, sending a beam of 3.60 x10^7 m/s protons (mass m) at a gas target of an unkno

matrenka [14]

Answer:

a) mass of unknown nucleus = 0.04245 mp, where mp is the proton mass

b) Speed of the unknown nucleus = (7.067 x 10^7) m/s

Explanation:

Considering the initial conditions, the observed collisions are ellastic, i.e, the total kinetic energy are conserved. The proton's mass will refer as $m_{p}$ .

(a)

Total kinetic energy conservation

$\frac{1}{2}m_{p}v_{p_0}^{2}+\frac{1}{2}m_{u}v_{u_o}^{2}=\frac{1}{2}m_{p}v_{p_f}^{2}+\frac{1}{2}m_{u}v_{u_f}^{2}$

where $v_{u_o}$ represents the initial velocity of the unknown element, $m_{u}$ the mass of the unknown element, and $v_{u_f}$ the final velocity of the unknown element

Linear momentum conservation

$m_{p}v_{p_0}+m_{u}v_{u_o}=m_{p}v_{p_f}+m_{u}v_{u_f}$

Using the initial speed of the target nucleus (unknown) is negligible, i.e, its speed is zero. Thereby, using the relation of linear momentum conservation given above, it is possible to find an expression of the final speed of the unknown nucleus in terms of its mass, which can be inserted in the relation of the kinetic energy conservation to obtain the value of the mass of the unknown elements, as follows;

$m_{u}v_{u_f}=m_{p}v_{p_0}-m_{p}v_{p_f}\\\\v_{u_f}=\frac{m_{p}(v_{p_0}-v_{p_f})}{m_{u}}$

Substituting this expression in the relation of total kinetic energy conservation,

$m_{p}(v^{2}_{p_0}-v^{2}_{p_f})={m_{u}}v^{2}_{u}_{f}$

Then,

$m_{p}(v^{2}_{p_0}-v^{2}_{p_f})={m_{u}}\frac{m^{2}_{p}(v_{p_0}-v_{p_f})^{2}}{m^{2}_{u}}\\\\m_{u}= \frac{m_{p}(v_{p_0}-v_{p_f})^{2}}{(v^{2}_{p_0}-v^{2}_{p_f})}$

Replacing the given data

$m_{u}= \frac{m_{p}(3.6x10^{7}-3.3x10^{7})^{2}}{((3.6x10^{7})^{2}-(3.3x10^{7})^{2})}$

Then,

$m_{u}=0.04245m_{p}$

(b) Using the relation of the final speed from linear momentum conservation and the above result, the speed of the unknown nucleus is calculated

$v_{u_f}=\frac{m_{p}(v_{p_0}-v_{p_f})}{m_{u}}\\\\v_{u_f}=\frac{m_{p}(3.6x10^{7}-3.3x10^{7})}{0.04245m_{p}}\\\\v_{u_f}=7.067x10^{7} m/s$

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3 years ago

If two overlapping waves have the same amplitude then what happens?

KonstantinChe [14]

I think the answer could be that it could result in resonance.

There is not much to go from in this question. Specify whether the are on the same wave length or on separate.

If the 2 waves are on the same wave length and going towards each other with the same amplitude, that would result in them colliding and forming a larger amplitude, or something called constructive interference.

I hope this helps you!

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3 years ago