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4 years ago

8

What are the first organisims to live within a newly created patch of land

2 answers:

horrorfan [7]4 years ago

8 0

The first organisms to live within a newly created patch of land are the Pioneer Species.

ss7ja [257]4 years ago

3 0

And the most common pioneer species are the lichens.

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Thermal Conductors don't have to be hot to transfer heat, explain a situation when a ice cube would still transfer heat to anoth

barxatty [35]

An ice cube would transfer heat to another object whose temperature
is lower than zero°C (32°F).

A block of "dry ice" is sitting there at a temperature of -78°C (-109°F).
An ice cube helps to melt dry ice nice and fast.

If you could find a block of solid nitrogen, its temperature would be
63K (-210°C, -346°F). An ice cube would transfer heat to that baby
so fast that it would instantly boil.

6 0

4 years ago

Uranium-235 decays to thorium-231 with a half-life of 700 million years. When a rock was formed, it contained 6400 million urani

Dahasolnce [82]

Answer:

proof in explanation

Explanation:

First, we will calculate the number of half-lives:

$n = \frac{t}{t_{1/2}}$

where,

n = no. of half-lives = ?

t = total time passed = 2100 million years

$t_{1/2}$ = half-life = 700 million years

Therefore,

$n = \frac{2100\ million\ years}{700\ million\ years}\\\\n = 3$

Now, we will calculate the number of uranium nuclei left ( $n_u$ ):

$n_u = \frac{1}{2^{n} }(total\ nuclei)\\\\n_u = \frac{1}{2^{3} }(6400\ million)\\\\n_u = \frac{1}{8}(6400\ million)\\\\n_u = 800\ million$

and the rest of the uranium nuclei will become thorium nuclei ( $u_{th}$ )

$n_{th} = total\ nuclei - n_u\\n_{th} = 6400\ million-800\ million\\n_{th} = 5600\ million$

dividing both:

$\frac{n_{th}}{n_u}=\frac{5600\ million}{800\ million} \\\\n_{th} = 7n_u$

<u>Hence, it is proven that after 2100 million years there are seven times more thorium nuclei than uranium nuclei in the rock.</u>

6 0

3 years ago

Gold in its pure form is too soft to be used for most jewelry. Therefore, the gold is mixed with other metals to produce an allo

Novay_Z [31]

Answer:

Explanation:18kt alloy contains

i) 75% of gold

rhogold=19.3g/cm^3

=75/100×19.3

=14.475g/cm^3

ii) 16% of silver

rhosilver=10.5g/cm^3

=16/100×10.5

=1.68g/cm^3

iii) 9% of copper

rhocopper =8.90g/cm^3

=9/100×8.9

=0.801g/cm^3

Overall density of 18kt gold

=(0.801+1.68+14.475)g/cm^3

=16.956g/cm^3

=17g/cm^3 to 3s.f

6 0

3 years ago

The electric current leaves the battery through the --- complete this blank

aev [14]

The bulb

Hope this helps!

5 0

4 years ago

Suppose that a steel bridge, 1000 m long, were built without any expansion joints. Suppose that only one end of the bridge was h

Stels [109]

Answer:

The difference in the length of the bridge is 0.42 m.

Explanation:

Given that,

Length = 1000 m

Winter temperature = 0°C

Summer temperature = 40°C

Coefficient of thermal expansion $\alpha= 10.5\times10^{-6}\ K^{-1}$

We need to calculate the difference in the length of the bridge

Using formula of the difference in the length

$\Delta L=L\alpha\Delta T$

Where, $\Delta T$ = temperature difference

$\alpha$ =Coefficient of thermal expansion

L= length

Put the value into the formula

$\Delta L=1000\times10.5\times10^{-6}(40^{\circ}-0^{\circ})$

$\Delta L=0.42\ m$

Hence, The difference in the length of the bridge is 0.42 m.

5 0

3 years ago