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poizon [28]
3 years ago
5

Sir charles sherrington observed that impulses took an unexpectedly long time to travel a neural pathway. his observation provid

ed evidence for the existence of
Biology
1 answer:
Karo-lina-s [1.5K]3 years ago
7 0

Sir Charles Sherrington’s observation has provided evidence for the existence of synaptic gaps in which this is the minute space provided in between of the target cell and the cell membrane of the axon terminal. This could be seen from his observation where the impulses took unexpectedly long time to travel a neural pathway.

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How does the release of trapped gases in magma cause a volcano to erupt?
dolphi86 [110]
D it causes lava to rush out faster
6 0
3 years ago
Read 2 more answers
A geologist is studying two different rhyolite flows to determine if they were erupted at the same time.
ss7ja [257]
Q1. The answer is 1.

It can be calculated using the equation:

 (1/2)ⁿ = x

x - decimal amount remaining, 

n - a number of half-lives.


x = 50% = 50/100 = 0.5

n = ?

(1/2)ⁿ = 0.5

log((1/2)ⁿ) = log(0.5)

n * log(1/2) = log(0.5)

n * log(0.5) = log(0.5)

n = log(0.5)/log(0.5)

n = 1


Q10. The answer is 2.


It can be calculated using the equation:

 (1/2)ⁿ = x

x - decimal amount remaining, 

n - a number of half-lives.


Rhyolite #2 has 25% of the parent H remaining:

x = 25% = 25/100 = 0.25

n = ?

(1/2)ⁿ = 0.25

log((1/2)ⁿ) = log(0.25)

n * log(1/2) = log(0.25)

n * log(0.5) = log(0.25)

n = log(0.25)/log(0.5)

n = -0.602 / - 0.301

n = 2



Q3. The answer is 100 million years.


A number of half-lives (n) is a quotient of total time elapsed (t) and length of half-life (H):

n = t/H

n = 1

t = ?

H = 100 000 000 years


n = t/H

t = n * H

t = 1 * 100 000 000 years

t = 100 000 000 years<span>

</span>

4 0
3 years ago
Chromosomes are made of <br> and contain two identical chromatids.
Verizon [17]

Answer:

these chromatids separate longitudinally to become individual chromosomes.

one chromosomes is composed of and DNA molecule

5 0
3 years ago
In a hardy-weinberg population with two alleles, a and a, that are in equilibrium, the frequency of allele a is 0.2. what is the
Allushta [10]
The Hardy-Weinberg equation is as follows:
( {p + q})^{2}  = 1
{p}^{2}  +2pq +   {q}^{2}  = 1
Where:
(convert all % to decimals)

p= homozygous dominant
q= homozygous recessive
pq= heterozygous

While you did not specify whether the 0.2 frequency was for dominant or recessive, we can still figure out the answer.

Using the 1st equation, we can solve for the other dominant/recessive frequency:

1-0.2=0.8

Meaning that:

p= 0.8 & q=0.2

If the heterozygouz frequency is 2pq, then it becomes a simple "plug & chug" sort of approach.

2(0.8)(0.2)= 2(0.16)= 0.32

So, the heterozygous frequency would be:
0.32

Hope this helps!
8 0
3 years ago
How can we test for HIV during the initial acute phase?
chubhunter [2.5K]
Acute HIV infection is the first phase of  HIV infection. The early HIV test can detect HIV during the acute HIV phase. The best test for detecting HIV infection is the early HIV test (also known as NAAT/RNA testing)

I Hope This Helped!
4 0
3 years ago
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