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Bingel [31]
2 years ago
9

The half life of uranium-235 is 4. 5 billion years. If 0. 5 half-lives have elapsed, how many years have gone by?

Physics
1 answer:
igor_vitrenko [27]2 years ago
5 0

Answer:

2.25 billion years

Explanation:

i am smort

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Ten narrow slits are equally spaced 3.00 mm apart and illuminated with indigo light of wavelength 444 nm. The width of bright fr
ankoles [38]

Answer:ask yo mama

Explanation:she finished school

8 0
3 years ago
If a star with an absolute magnitude of -5 has an apparent magnitude of +5 ,then its distance is
klio [65]
You asked a question.  I'm about to answer it. 
Sadly, I can almost guarantee that you won't understand the solution. 
This realization grieves me, but there is little I can do to change it. 
My explanation will be the best of which I'm capable.


Here are the Physics facts I'll use in the solution:

-- "Apparent magnitude" means how bright the star appears to us.

-- "Absolute magnitude" means the how bright the star WOULD appear
if it were located 32.6 light years from us (10 parsecs).

-- A change of 5 magnitudes means a 100 times change in brightness,
so each magnitude means brightness is multiplied or divided by  ⁵√100 .
That's about  2.512... .  

-- Increasing magnitude means dimmer.
Decreasing magnitude means brighter.
+5 is 10 magnitudes dimmer than -5 .

-- Apparent brightness is inversely proportional to the square
of the distance from the source (just like gravity, sound, and
the force between charges).

That's all the Physics.  The rest of the solution is just arithmetic.
____________________________________________________

-- The star in the question would appear M(-5) at a distance of
32.6 light years. 

-- It actually appears as a M(+5).  That's 10 magnitudes dimmer than M(-5),
because of being farther away than 32.6 light years.

-- 10 magnitudes dimmer is ( ⁵√100)⁻¹⁰ = (100)^(-2) .

-- But brightness varies as the inverse square of distance,
so that exponent is (negative double) the ratio of the distances,
and the actual distance to the star is

(32.6) · (100)^(1) light years

= (32.6) · (100) light years

=  approx.  3,260 light years .   (roughly 1,000 parsecs)


I'll have to confess that I haven't done one of these calculations
in over 50 years, and I'm not really that confident in my result.
If somebody's health or safety depended on it, or the success of
a space mission, then I'd be strongly recommending that you get
a second opinion.
But, quite frankly, I do feel that mine is worth the 5 points.
6 0
2 years ago
A dandelion seed floats to the ground in a mild wind with a resultant velocity of 26.0 cm/s. If the horizontal component velocit
stira [4]

Answer:

24 cm/s

Explanation:

Applying

Pythagoras theorem,

a² = b²+c²............. Equation 1

Where a = resultant, b = vertical component, c = horizontal component

From the question,

Given: a = 26 cm/s, c = 10 cm/s

Substitute these values into equation 1

26² = b²+10²

676 = b²+100

b² = 676-100

b² = 576

b = √576

b = 24 cm/s

7 0
2 years ago
A galaxy that is a featureless spherical ball of stars would be called a type
satela [25.4K]

Answer:

E0

Explanation:

Yes. The "0" indicates that it is spherical

7 0
2 years ago
A pen cap floats in a plastic lemonade 3/4 full of water vif you squeeze the bottle the pen cap sinks to the bottom if you then
elena-s [515]

Answer:

The pressure of the air molecules inside the pen cap increases and the volume occupied by the air decreases such that the combined volume occupied by the pen cap and the air volume reduces while the combined mass of the pen cap and the air molecules remain the same

Given that density = The mass/Volume, we have that the density varies inversely as the volume, and as the volume reduces, the density increases

Upon squeezing, therefore, as the new combined density of the pen cap and the air molecules rises to more than the density of the water in the bottle, then, the pen cap air molecule is relatively more denser than the water, which will result in the pen cap sinking to the bottom of the bottle

Explanation:

4 0
3 years ago
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