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SCORPION-xisa [38]
3 years ago
14

How did Linnaeus improve the classification of organisms

Physics
2 answers:
mamaluj [8]3 years ago
6 0

Answer: A

Reason: one of his contributions was the development of a hierarchical system of classification of nature. Linnaeus also provided us with a consistent way to name species called binomial nomenclature. (From Internet)

Darina [25.2K]3 years ago
5 0

Answer: my screen looks exactly the same and i was looking for the answer swell but i noticed that the class matches mines as well,im trying to say i think that we know each other plz reply if you go to Georgia jones ayers middle school, and have  teacher name ms.banks

Explanation:

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Manny walked a total of 3 miles. The reference point used to calculate the total distance that he walked was the same as the end
yawa3891 [41]

Correct question:

Manny walked a total of 3 miles. The reference point used to calculate the total distance that he walked was the same as the ending point. Which describes where Manny most likely walked?

a. from the bottom of a hill to the top

b. on a circular nature trail

c. on a sidewalk from his house to the mall

d. from the beginning of a straight track to the end

Answer:

b. on a circular nature trail

Explanation:

As it is mentioned that Manny that his reference point from where she started is same as ending point meaning that she moved in a circular path making point B correct answer.

7 0
3 years ago
Read 2 more answers
A sodium ion, Na+ , has a positive charge because it
goldenfox [79]

D. lost an electron

because they have free electron in the outermost shell which they will like to denote, in order to attain stable configuration

3 0
3 years ago
If you drive 125 km in 2 hours, what is your average speed
erma4kov [3.2K]
Try Socratic it helped me with this type of question
7 0
3 years ago
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The density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cubic
Drupady [299]

a) Density at 100 degrees: 1.34\cdot 10^4 kg/m^3

Explanation:

The density of mercury at 0 degrees is d=1.36\cdot 10^4 kg/m^3

Let's take 1 kg of mercury. Its volume at 0 degrees is

V=\frac{m}{d}=\frac{1 kg}{1.36\cdot 10^4 kg/m^3}=7.35\cdot 10^{-5} m^3

The formula to calculate the volumetric expansion of the mercury is:

\Delta V= \alpha V \Delta T

where

\alpha=180\cdot 10^{-6} K^{-1} is the cubic expansivity of mercury

V is the initial volume

\Delta T is the increase in temperature

In this part of the problem, \Delta T=100 C-0 C=100 C=100 K

So, the expansion is

\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(100 K)=1.3\cdot 10^{-6} m^3

So, the new density is

d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+1.3\cdot 10^{-6} m^3}=1.34\cdot 10^4 kg/m^3


b) Density at 22 degrees: 1.355\cdot 10^4 kg/m^3

We can apply the same formula we used before, the only difference here is that the increase in temperature is

\Delta T=22 C-0 C=22 C=22 K

And the volumetric expansion is

\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(22 K)=2.9\cdot 10^{-7} m^3

So, the new density is

d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+2.9\cdot 10^{-7} m^3}=1.355\cdot 10^4 kg/m^3


8 0
3 years ago
I need help with this answer
Gemiola [76]

decomposition

A decomposition reaction is just the opposite of combination reaction

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3 years ago
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