In preparing to shoot an arrow, an archer pulls a bow string back 0.424 m by exerting a force that increases uniformly from 0 to
1 answer:
Answer:
525.94N/m
Explanation:
According to Hooke's law, the extension or compression of an elastic material is proportional to an applied force provided that the elastic limit of the material is not exceeded.
where F is the applied force or load, k is the elastic constant or stiffness of the material and e is the extension.
In this problem, the bow string is assumed to behave like an elastic material that is stretched not beyond the elastic limit.
Given;
F = 223N;
e = 0.424m
k = ?
We make substitutions into equation (1) and then solve for k.
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Answer:
Si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.
Explanation:
Supongamos que el vehículo de Speedy Sue decelera a razón constante, mientras que la camioneta se desplaza a velocidad constante. Se requiere conocer si ambos vehículos colisionarán, lo cual implica conocer si existe algún instante tal que ambos tengan la misma posición. Consideremos además que la posición de referencia se encuentra en la posición inicial de Sppedy Sue. Entonces, las ecuaciones cinemáticas son:
Speedy Sue
Camioneta lenta
Donde:
,
- Posiciones iniciales de Speedy Sue y la camioneta lenta, medidas en metros.
- Velocidad inicial de Speedy Sue, medida en metros por segundo.
- Velocidades actuales de Speedy Sue y la camioneta lenta, medidas en metros por segundo.
- Deceleración de Speedy Sue, medida en metros por segundo cuadrado.
- Tiempo, medido en segundos.
Si conocemos que ,
,
,
,
y
, encontramos la siguiente función cuadrática:
(Ec. 1)
Las raíces de esta función son:
,
La colisión ocurriría en la raíz positiva más pequeña, es decir:
Ahora, la posición en que ocurriría la colisión se determina a partir de la ecuación de desplazamiento de la camioneta lenta, es decir: (,
,
)
En síntesis, si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.
Answer: Choice B
There are triple bonds between the carbon (C) and oxygen (O) atoms. Then there are 2 dots on either side
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Explanation:
When it comes to interaction and chemistry, all that matters is the valence shell or valence electrons. This is the outermost shell. This is because various elements do not interact with the inner electrons (they're locked in place so to speak and don't move to other elements).
Carbon has 6 protons, which is what uniquely makes up this element. This means there are 6 electrons. The inner shell has 2 electrons and the valence shell has 4 electrons. Two electrons are shown as the two blue dots on the left side of the C. The other two electrons form two of the lines, or the bonds, between the C and O.
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Oxygen has 8 protons and 8 electrons. It has 2 electrons in the inner shell and 6 electrons in the valence shell. Two of those electrons are the red dots on the right side of the O. The other 4 electrons are shared to form the bonds with the carbon atom.
This is where things get a bit tricky. I've shown a diagram below indicating that one of the oxygen electrons (red dot) is passed to the carbon, as this carbon atom is pulling on the oxygen electron. But the oxygen atom is pulling on it as well, which forms one of the triple bonds.
So this is why diagram B is the final answer. This is something you can logically determine (remembering the rules of how each electron shell is formed), or it's something you'll need to memorize. In the real world, it's easy to forget a lot of info like this, so that's why having it handy as a lookup reference is preferable.
