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3 years ago

Not sure what to put this under

Physics

1 answer:

cestrela7 [59]3 years ago

5 0

family 16 cause i said so XD

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imagine you are going on a rid in a spacecraft next to earth. Your trip takes one whole year. Describe earth's tilt in the north

mylen [45]

You will have to fly around the whole earth to get to your landing station

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3 years ago

Can Someone please help me! <br><br> What is deposition

7nadin3 [17]

Deposition is the geological process in which sediments, soil and rocks are added to a landform or landmass

7 0

3 years ago

Read 2 more answers

An energy storage system based on a flywheel (a rotating disk) can store a maximum of 3.7 MJ when the flywheel is rotating at 16

Likurg_2 [28]

The moment of inertia of the flywheel is 2.63 kg- $m^{2}$

It is given that,

The maximum energy stored on the flywheel is given as

E=3.7MJ= 3.7× $10^{6}$ J

Angular velocity of the flywheel is 16000 $\frac{rev}{min}$ = 1675.51 $\frac{rad}{sec}$

So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

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

By rearranging the equation:

I = $\frac{2E}{w_{2} }$

I = 2.63 kg- $m^{2}$

Thus the moment of inertia of the flywheel is 2.63 kg- $m^{2}$ .

Learn more about moment of inertia here;

brainly.com/question/13449336

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7 0

1 year ago

How does friction make it possible for you to walk across the floor?

Tema [17]

if we are walking on a perfectly smooth ground which has no friction our force would simply cancel out the force reverted by the ground and we would fall.

We need it to help push out feet off the ground

Hope those helps :)

5 0

3 years ago

Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)

sergeinik [125]

$v^2-{v_0}^2=2a(x-x_0)$

dónde $v$ es la velocidad de la esfera, $v_0$ es suya velocidad inicial, $a=-g$ la aceleración debida a la gravedad, $x$ la posición, y $x_0$ la posición inicial. Tomamos $x_0=0\,\mathrm m$ a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

$v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)$

$\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}$

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

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3 years ago