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3 years ago

If all the stars in an elliptical galaxy traveled random directions in their orbits, the elliptical galaxy would be type

1 answer:

7 0

The answer would be E7. Galaxies categorized as E0 look to be nearly perfect, while those registered as E7 seem much extended than they are widespread. It is worth noting, though, that a galaxy's look is connected to how it lies on the sky when viewed from Earth. An E7 galaxy is very long and thin or the flattest of them all.

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Work is being done in which of these situations? All motions are at a constant velocity (including velocity = zero).

Black_prince [1.1K]

Work is done only when there is a displacement of the object on which a force is applied in the direction of the force.

4 0

3 years ago

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How do earths magnetic pole reversals provide evidence for plate tectonics

Alinara [238K]

Rocks get older as you move away from the water surface

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3 years ago

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The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

6 0

4 years ago

The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years. Convert this distance

alukav5142 [94]

Explanation:

The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years.

Light year is the unit of distance covered by the heavenly bodies. 1 light year is equal to :

$1\ light\ year=3.72\times 10^{17}\ inches$

So, $4.218\ light\ year=4.218\times 3.72\times 10^{17}\ inches$

$4.218\ light\ year=1.56\times 10^{18}\ inches$

We need to convert 4.218 light-years barley corns.

Since, 1 barleycorn = 1/3 inch

$1\ inch=3\ barleycorn$

$1.56\times 10^{18}\ inches=3\times 1.56\times 10^{18}=4.68\times 10^{18}\ barleycorn$

So, the nearest star to the Earth is at a distance of $4.68\times 10^{18}\ barleycorn$ . Hence, this is the required solution.

3 0

3 years ago

Camina con otro compañero, al mismo tiempo y al mismo paso. Quién se mueve tú o tu compañero? Razona tu respuesta

Marrrta [24]

Answer:

Fricción.

Explicación:

Caminando con otro compañero al mismo tiempo y al mismo ritmo, el movimiento se produce debido a la fricción del suelo y la suela de los zapatos porque la fricción es la fuerza que ayuda en el movimiento de los objetos de un lugar a otro. Si no hay fricción entre el suelo y la suela de los zapatos, no podemos dar un paso por lo que podemos decir que la fricción nos mueve hacia adelante.

5 0

3 years ago