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3 years ago

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1- ¿Cuál sería la energía de un objeto de 50 newton de peso que se encuentra sobre una estantería de 3 metros de altura? ¿Qué ti

po de Energía es?

1 answer:

Liula [17]3 years ago

6 0

Responder:

<h3>150 Nm </h3><h3>Energía potencial </h3>

Explicación:

El tipo de energía que posee el objeto se conoce como energía potencial. <u>La energía potencial es la energía que posee un objeto, mi virtud de su posición. </u>

Energía potencial = masa * aceleración debido a la gravedad * altura

Dado que Force = masa * aceleración debido a la gravedad

Energía potencial = Fuerza * altura

Fuerza dada = 50N y altura = 3 m

Energía potencial = 50 * 3

Energía potencial = 150 Nm

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The 1.5kg ball is launched straight upward with an initial velocity of 7m/s. What is the maximum height it will reach?

Temka [501]

Answer:

h = 2.5 m

Explanation:

Given that,

Mass of a ball, m = 1.5 kg

Initial velocity of the ball, u = 7 m/s

We need to find the maximum height reached by the ball. Let it is be h. Using the conservation of energy to find it such that,

$mgh=\dfrac{1}{2}mu^2\\\\h=\dfrac{u^2}{2g}$

Put all the values,

$h=\dfrac{7^2}{2\times 9.8}\\\\=2.5\ m$

So, it will reach to a height of 2.5 m.

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3 years ago

What are some differences you can spot in a periodic table?

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3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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natali 33 [55]

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3 years ago