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2 years ago

12

todos los cuerpos pueden estar sobre la tierra por la fuerza de gravedad .cuando nos caemos ,saltamos y realizamos otras activid

ades la fuerza de gravedad influye bastante ,por lo tanto ,esta fuerza la ejerce

1 answer:

const2013 [10]2 years ago

4 0

¿Es esta una pregunta verdadera o falsa? Por cierto, no hay mucha gente que hable español en esta aplicación, así que buena suerte.

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You have a spring-loaded air rifle. When it is loaded, the spring is compressed 0.3 m and has a spring constant of 150 N/m. In j

Feliz [49]

The potential energy of the spring is 6.75 J

The elastic potential energy stored in the spring is given by the equation:

$E= \frac{1}{2} kx^2$

where;

k is the spring constant

x is the compression/stretching of the string

In this problem, we have the spring as follows:

k = 150 N/m is the spring constant

x = 0.3 m is the compression

Substituting in the equation, we get

$E=\frac{1}{2} (150) (0.3)^2$

$E=6.75J$

Therefore. the elastic potential energy stored in the spring is 6.75J .

Learn more about potential energy here:

brainly.com/question/10770261

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1 year ago

What is the velocity of a wave with a wavelength of 3.7 m and a frequency of 22.5 Hz?

ExtremeBDS [4]

Answer: 343 m/s. The sound wave has a frequency of. 436 Hz. What is the period of the wave? T = = 436 Hz. = 2.29x10-3 s. C. What is the wave's wavelength? To halve

Explanation:

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2 years ago

An atom that has fewer neutrons than protons and more electrons than protons is a ion

riadik2000 [5.3K]

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3 years ago

Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa

lyudmila [28]

Answer:

No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field $\vec{E}$ is related to the gradient of the electric potential V with :

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})$

This means that for constant electric potential the electric field must be zero:

$V(\vec{r}) = k$

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k$

$\vec{E} (\vec{r}) = - (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k$

$\vec{E} (\vec{r}) = - (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})$

$\vec{E} (\vec{r}) = - (0,0,0)$

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

$V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4$

give an electric field of zero at point (0,0,0)

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3 years ago

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kiruha [24]

I’m so sorry, I need more information. Good luck and I’m sorry I couldn’t help you :(

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3 years ago