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3 years ago

What is the purpose of the traveling gnome experiment? (Search up traveling gnome experiment)

1 answer:

7 0

After conducting in-depth research into education and science influencers - delving into scientific theories on weight and measurement - we conceived of the Gnome Experiment – a global research project aimed at proving the scientific theory that gravity varies from place to place affecting weight.

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In spiral galaxies, the relationship between nuclear bulge and tightness of spiral arms seems to be that

Rus_ich [418]

In spiral galaxies, the relationship between nuclear bulge and tightness of spiral arms seems to be that the larger the nuclear bulge of the galaxy, tighter the spiral arms are wound together.

What are spiral galaxies?

Spiral galaxies form a class of galaxy originally described by Edwin Hubble in his 1936 work The Realm of the Nebulae and, as such, form part of the Hubble sequence.

Most spiral galaxies consist of a flat, rotating disk containing stars, gas and dust, and a central concentration of stars known as the bulge. These are often surrounded by a much fainter halo of stars, many of which reside in globular clusters.

Spiral galaxies are named by their spiral structures that extend from the center into the galactic disc.

The spiral arms are sites of ongoing star formation and are brighter than the surrounding disc because of the young, hot OB stars that inhabit them.

Roughly two-thirds of all spirals are observed to have an additional component in the form of a bar-like structure, extending from the central bulge, at the ends of which the spiral arms begin.

The proportion of barred spirals relative to bar less spirals has likely changed over the history of the universe, with only about 10% containing bars about 8 billion years ago, to roughly a quarter 2.5 billion years ago, until present, where over two-thirds of the galaxies in the visible universe (Hubble volume) have bars.

To learn more about spiral galaxies: brainly.com/question/14243370

#SPJ4

7 0

1 year ago

Consider the concepts of kinetic energy (KE) and gravitational potential energy (GPE) as you complete these questions. A ball is

Lena [83]

Answer:

When the ball is held motionless above the floor, the ball possesses only GPE energy.If the ball is dropped, its GPE energy decreases as it falls.If the ball is dropped, its KE energy increases as it falls.

Explanation:

If the ball is held motionless, then its kinetic energy is equal to zero, since kinetic energy depends on the velocity. And the ball is held above the ground, which means it possesses gravitational potential energy.

If the ball is dropped, its height will decrease, therefore its gravitational potential energy will decrease. Along the way, the ball will be in free fall, and therefore its velocity will increase, hence its kinetic energy.

$K = \frac{1}{2}mv^2\\U = mgh$

3 0

3 years ago

A bowling ball rolls 33\,\text m33m33, start text, m, end text with an average speed of 2.5\,\dfrac{\text m}{\text s}2.5 s m 2

asambeis [7]

The ball rolled for 13.2 s

<h3>Further explanation</h3>

Speed is scalar and no direction

$\tt avg~speed=\dfrac{total~distance}{elapsed~time}=\dfrac{\Delta x}{\Delta t}$

A bowling ball rolls 33 m, with average speed = 2.5 m/s

So elapsed time :

$\tt t=\dfrac{33~m}{2.5`m/s}=13.2~s$

4 0

3 years ago

A ball of radius R and mass m is magically put inside a thin shell of the same mass and radius 2R. The system is at rest on a ho

dlinn [17]

Answer:

$x =\frac{-R}{2}$

Explanation:

From the question we are told that mass

Thin layer radius $= 2R$

Generally the expression for ths solution is given as

Xcm =(m*0 =m(-2R))/2m =-mR/(2m)=-R/2

the center of mass will not move at initial state

Considering the center of mass of both bodies

$xcm=\frac{m*x+m*x)}{2m} =x$

$x =\frac{-R}{2}$

Therefore the enclosing layer moves $x =\frac{-R}{2}$

7 0

3 years ago

A resistor with an unknown resistance is connected in parallel to a 13 Ω resistor. When both resistors are connected in parallel

larisa86 [58]

Answer:

R2 = 10.31Ω

Explanation:

For two resistors in parallel you have that the equivalent resistance is:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\\$ (1)

R1 = 13 Ω

R2 = ?

The equivalent resistance of the circuit can also be calculated by using the Ohm's law:

$I=\frac{V}{R_{eq}}\\\\R_{eq}=\frac{V}{I}$ (2)

V: emf source voltage = 23 V

I: current = 4 A

You calculate the Req by using the equation (2):

$R_{eq}=\frac{23V}{4A}=5.75\Omega$

Now, you can calculate the unknown resistor R2 by using the equation (1):

$\frac{1}{R_2}=\frac{1}{R_{eq}}-\frac{1}{R_1}\\\\R_2=\frac{R_{eq}R_1}{R_1-R_{eq}}\\\\R_2=\frac{(5.75\Omega)(13\Omega)}{13\Omega-5.75\Omega}=10.31\Omega$

hence, the resistance of the unknown resistor is 10.31Ω

8 0

3 years ago

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