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3 years ago

Most of the bright stars in our galaxy are located in the galactic

Physics

1 answer:

GenaCL600 [577]3 years ago

3 0

False. Most of the bright stars in our galaxy are located in the disk.

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if a star 100 light years from earth is beginning to expand into a giant star how long will it take for astronomers to observe t

VikaD [51]

Answer:

100years later

Explanation:

Because the lights will arrive at world after 100 years later.

7 0

3 years ago

In an experiment you are performing, your lab partner has measured the distance a cart has traveled: 28.4inch. You need the dist

weeeeeb [17]

Answer: The multiplication factor is 72.136 cm. This will give you the unit conversion when multiplied with 28.4 inch

Explanation:

1 inch = 2.54 cm

28.4 inches = x cm

Xcm= (28.4 inches × 2.54cm)/1 inch

X= 72.136

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3 years ago

A scene in a movie has a stuntman falling through a floor onto a bed in the room below. The plan is to have the actor fall on hi

IgorC [24]

Answer:

$m=17.79Kg$

Explanation:

In this process energy must be conserved. On the initial stage, there will be only gravitational potential energy, while on the final stage there will be only elastic potential energy, so they will be equal. We write this as:

$U_g=U_e$

Which is the same as:

$mgh=\frac{k \Delta x^2}{2}$

So we can obtain our mass from there, and for our values:

$m=\frac{k \Delta x^2}{2gh}=\frac{(65144 N/m)(0.1333m)^2}{2(9.8m/s^2)(3.32m)}=17.79Kg$

4 0

3 years ago

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

$v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t$

$v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}$

At <em>t</em> = 5.0 s, the particle has velocity

$v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}$

$v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}$

The speed at this time is the magnitude of the velocity :

$\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}$

The direction of motion at this time is the angle $\theta$ that the velocity vector makes with the positive <em>x</em>-axis, such that

$\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}$

4 0

2 years ago

Why does it take double the applied force to move a mass double the size?

Ratling [72]

56-999999999999999999999-4 is the best for my mom

5 0

3 years ago