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2 years ago

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Una persona pesa 700 N al nivel del mar. ¿Cuándo sube a un avión a 10000 m de altura pesará más, igual o menos?

1 answer:

ASHA 777 [7]2 years ago

8 0

Answer:

La persona pesa menos

Explanation:

Generalmente, definimos el peso como la fuerza gravitacional que actúa sobre un objeto.

Dado que la constante gravitacional depende del cuadrado de la distancia medida desde el centro de la Tierra, podemos concluir que el peso de un objeto disminuye a medida que aumenta la distancia entre el objeto y el centro de la Tierra (mayor altitud).

Por lo tanto, la persona que pesa 700 N al nivel del mar pesará menos cuando esté en un avión a una altitud de 10,000 m.

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Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write

Eva8 [605]

Answer:

R = ( 4.831 m , 1.469 m )

Magnitude of R = 5.049 m

Direction of R relative to the x axis= 16°54'33'

Explanation:

Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

$\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ its the magnitude and θ.

So, for our vectors, we will have:

$\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )$

$\vec{D}= ( 2.121 m , -2.121 m )$

and

$\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )$

$\vec{E}= ( 2.71 m , 3.59 m )$

Now, we can take the sum of the vectors

$\vec{R} = \vec{D} + \vec{E}$

$\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )$

$\vec{R} = ( 2.121 \ m + 2.71 \ m , -2.121 \ m + 3.59 \ m )$

$\vec{R} = ( 4.831 \ m , 1.469 \ m )$

This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}$

$|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}$

$|\vec{R}| = \sqrt{25.496 m^2}$

$|\vec{R}| = 5.049 m$

To find the direction, we can use

$\theta = arctan(\frac{R_y}{R_x})$

$\theta = arctan(\frac{1.469 \ m}{4.831 \ m})$

$\theta = arctan(0.304)$

$\theta = 16\°54'33''$

As we are in the first quadrant, this is relative to the x axis.

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Monochromatic light of wavelength λ=136.8μ m is shone at normal incidence through a thin film of thickness t resting atop a full

inn [45]

Answer:

1.8 × 10⁻⁸ Hm

Explanation:

Given that:

The refractive index of the film = 19

The wavelength of the light = 136.8 μ m

The thickness can be calculated by using the formula shown below as:

$Thickness=\frac {\lambda}{4\times n}$

Where, n is the refractive index of the film

${\lambda}$ is the wavelength

So, thickness is:

$Thickness=\frac {136.8\ \mu\ m}{4\times 19}$

Thickness = 1.8 μ m

Since,

1 μ m = 10⁻⁸ Hm

So,

Thickness = 1.8 × 10⁻⁸ Hm

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2 years ago