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Tanzania [10]
3 years ago
11

You observe a spiral galaxy with a large central bulge and tightly wrapped arms. It would be classified a

Physics
1 answer:
miv72 [106K]3 years ago
6 0

Answer:

Sa

Explanation:

Spiral Galaxies  -

It is a disk shaped galaxies which have spiral structure , is refereed to as spiral galaxies .

According to Hubble , these galaxies are classified as Sa , Sb , Sc .

Where ,

Sa - have the structure , which is bulged from the central portion , along with a tightly wrapped spiral structure .

Sb - have a lesser bulge and the spiral is looser .

Sc - It has very weak bulge with the open spiral structure .

Hence , from the question ,

The given information is about the Sa .

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You input 75 J of work with a wedge. If the wedge does 65 J of useful work, what if the efficiency of the wedge ?
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Answer:

Efficiency of wedge is the ratio of "work done by the machine to the work supplied to the machine".

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A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m ab
maria [59]

Answer:

W = 0.678 rad/s  

Explanation:

Using the conservation of energy:

E_i =E_f

Roll up and hill without slipping is the sumatory of two energys, rotational and translational, so:

\frac{1}{2}IW^2+ \frac{1}{2}mV^2 = mgh

where I is the moment of inertia, W the angular velocity at the base of the hill, m the mass of the ball, V the velocity at the base of the hill, g the gravity and h the altitude.

First, we will find the moment of inertia as:

I =\frac{2}{3}mR^2

where m is the mass and R the radius, so:

I =\frac{2}{3}(0.426kg)(11.3m)^2

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Then, replacing values on the initial equation, we get:

\frac{1}{2}(36.26)W^2+ \frac{1}{2}(0.426kg)V^2 = (0.426kg)(9.8)(5m)

also we know that:

V =WR

so:

\frac{1}{2}(36.26)W^2+ \frac{1}{2}(0.426kg)W^2R^2 = (0.426kg)(9.8)(5m)

Finally, solving for W, we get:

W^2(\frac{1}{2}(36.26)+ \frac{1}{2}(0.426kg)(11.3m)^2) = (0.426kg)(9.8)(5m)

W = 0.678 rad/s

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3 years ago
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