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OverLord2011 [107]
3 years ago
8

An atom is the most __________ unit of living and nonliving things.

Physics
2 answers:
storchak [24]3 years ago
7 0
Basic is the answer. (i think so) 
Bingel [31]3 years ago
3 0
Basic is the answer!
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When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these
Zigmanuir [339]

Answer:

a)  v = 0.7071 v₀, b) v= v₀, c)  v = 0.577 v₀, d)   v = 1.41 v₀, e)  v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

          v = \sqrt{ \frac{T}{ \mu } }

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

          m = 2m₀

let's find the new linear density

          μ = m / l

iinitial density

          μ₀ = m₀ / l

final density

          μ = 2m₀ / lo

          μ = 2 μ₀

we substitute in the equation for the velocity

initial            v₀ = \sqrt{ \frac{T_o}{ \mu_o} }

with the new dough

                    v = \sqrt{ \frac{T_o}{ 2 \mu_o} }

                    v = 1 /√2  \sqrt{ \frac{T_o}{ \mu_o} }

                    v = 1 /√2 v₀

                    v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

              μ = μ₀

   in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

           μ = 3m₀ / l₀

           μ = 3 m₀/l₀

           μ = 3 μ₀

            v = \sqrt{ \frac{T_o}{ 3 \mu_o} }

            v = 1 /√3  v₀

            v = 0.577 v₀

d) l = 2l₀

            μ = m₀ / 2l₀

            μ = μ₀/ 2

           v = \sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }

           v = √2 v₀

            v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

             μ = 10 m₀/2l₀

             μ = 5 μ₀

we look for speed

             v = \sqrt{ \frac{T_o}{5 \mu_o} }

             v = 1 /√5  v₀

             v = 0.447 v₀

5 0
3 years ago
Explain what happens during stages A and B.
fgiga [73]
Is there a picture?
3 0
3 years ago
When a red giant dies, before it becomes a WHITE DWARF (small, dim, hot star) that eventually becomes a BLACK DWARF (remnant, da
vagabundo [1.1K]

some massive black dwarfs may eventually produce <u>supernova explosions. </u>These will occur if pycnonuclear (density-based) fusion processes much of the star to iron, which would lower the Chandrasekhar limit for some black dwarfs below their actual mass.

6 0
3 years ago
Based on your calculations in Activity 3.7B, what are the rates of seafloor spreading for the North Atlantic basin ______cm/yr
Over [174]
It was be at least 67 cm/yr for the North Atlantic
3 0
3 years ago
For an unknown sample of the experiment, students measure 1679 counts when they first receive their sample and 1336 counts four
Tpy6a [65]

Answer:

Half life of the sample, t_{\frac{1}{2}} = 12.15 min

Given:

Initial amount, N = 1679

Final count of amount, N' = 1336

Time elapsed, t = 4 min = 240 s

Solution:

Now, To calculate the half life, using the relation:

N' = N(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

Now, substituting the given values in the above mentioned formula:

\frac{1336}{1679} = (\frac{1}{2})^{\frac{4}{t_{\frac{1}{2}}}}

0.796 = 0.5^{\frac{4}{t_{\frac{1}{2}}}}

Taking log on both the sides:

ln(0.796) = \frac{4}{t_{\frac{1}{2}}}ln(0.5)

0.329 = 4t_{\frac{1}{2}}

t_{\frac{1}{2}} = 12.15 min

6 0
3 years ago
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