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3 years ago

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An object at rest stays at rest and an object in stays in motion with the same speed and in the same direction unless acted upon

by an unbalanced force is Newton’s 1st law of motion
True or false

1 answer:

IrinaVladis [17]3 years ago

3 0

Answer:

Hi I'm happy to meet you and the answer for your question is true. If you need me to state the law ask me

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What is the value of work done on an object when a 70-newton force moves it 9.0 meters in the same direction as the force?

marta [7]

630 watts - 9 x 70 = 630

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4 years ago

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If a woman runs 100 meters north and then 70 meters south, what is her total displacement?

Burka [1]

Displacement is the distance between the starting and ending points, in the direction from the starting point to the ending point, all regardless of the route followed from the starting point to the end point.

The woman's displacement is therefore 30 meters north.

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Chapter 38, Problem 001

Norma-Jean [14]

Answer:

a) $\lambda=2.95x10^{-6}m$

b) infrared region

Explanation:

Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".

Part a

Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:

$E=hf$

Where h is the Planck's Constant. By the other hand the know that $c=f\lambda$ and if we solve for f we have:

$f=\frac{c}{\lambda}$

If we replace the last equation into the E formula we got:

$E=h\frac{c}{\lambda}$

And if we solve for $\lambda$ we got:

$\lambda =\frac{hc}{E}$

Using the value of the constant $h=4.136x10^{-15} eVs$ we have this:

$\lambda=\frac{4.136x10^{15}eVs (3x10^8 \frac{m}{s})}{0.42eV}=2.95x10^{-6}m$

$\lambda=2.95x10^{-6}m$

Part b

If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of $10^{-6}m$

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3 years ago

Why was the concern over global cooling replaced with a concern over global warming?

velikii [3]

<span>During 1970s, same observations were seen as what we have observed today pertaining to our climate. Journals were discussing that there would be warming because of greenhouse gases emissions. Also, it was observed between the years 1970 to 1990 that there was a steady surface temperature increase. Due to this, people are now fixated with global warming rather than on global cooling.</span>

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3 years ago

Como anticipan la caída del proyectil, con movimiento parabólico para que el misil no haga daño

Elena L [17]

Answer:

Conociendo la velocidad inicial del proyectil y el angulo de lanzamiento con respecto ala horizontal.

Explanation:

Para poder anticipar la caída del proyectil es importante conocer la velocidad inicial del proyectil y el angulo de disparo del proyectil con respecto a la horizontal.

A continuación se presenta un diagrama o esquema donde se pueden ver estas variables y se explicaran a la brevedad:

Para poder encontrar el rango que es la máxima distancia horizontal recorrida por el proyectil debemos utilizar la siguiente ecuación:

$x=(v_{o})_{x} *t\\where:\\(v_{o})_{x} = velocidad inicial x-component [m/s]\\t= time [s]$

Para poder encontrar el tiempo debemos utilizar la siguiente ecuación:

$y=(v_{y} )_{o}*t-0.5*g*t^{2} \\donde:\\(v_{y} )_{o}= velocidad inicial componente y [m/s]\\g = gravity = 9.81 [m/s^2]\\t = time [s]$

En la anterior ecuación, igualamos y = 0, ya que cuando el proyectil cae al suelo la distancia vertical es cero. De esta manera podemos encontrar el tiempo t, ya que conocemos la velocidad inicial del proyectil en la componente y.

Seguidamente reemplazamos t en la primera ecuacion y encontramos la distancia x o el rango.

8 0

3 years ago