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3 years ago

7

What r three positive results of a variation within a population that occur due to natural selection

1 answer:

Phoenix [80]3 years ago

4 0

Answer:

Traits, evolution, adaptive

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A car is moving at 34mi/hr. If the car travels for 6hours how many miles (mi) did the car travel? Make sure you include the prop

Dmitrij [34]

the car travels 34 mi in one hour.

then, in 6 hours car travels

34 x 6 mi

= 204 mi

5 0

3 years ago

Read 2 more answers

Why was the idea of plate tectonics difficult for many scientists to accept for many years after it was first introduced?

pentagon [3]

I think the correct answer from the choices listed above is the second option. The <span> idea of plate tectonics was difficult for many scientists to accept for many years after it was first introduced because there </span><span>was no explanation yet for how it was happening. It was only to the recent times that these were proven. </span>

3 0

3 years ago

Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy

8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

Formula used for the radius of the $n^{th}$ orbit will be,

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

$E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV$

Energy of the second orbit in H atom .

$E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV$

Energy of the third orbit in H atom .

$E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV$

Energy of the fifth orbit in H atom .

$E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV$

Energy of the sixth orbit in H atom .

$E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV$

Energy of the seventh orbit in H atom .

$E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV$

During an absorption of energy electron jumps from lower state to higher state.So, absorption will take place in :

1) n = 2 → n = 3

2) n= 5 → n = 6

Energy absorbed when: n = 2 → n = 3

$E=E_3-E_2$

$E=(-1.51 eV) -(-3.40 eV)=1.89 eV$

Energy absorbed when: n = 5 → n = 6

$E'=E_6-E_5$

$E'=(-0.378 eV)-(-0.544 eV) =0.166 eV$

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0

3 years ago

Una banda transportadora se utiliza para cargar canastos en un avión de carga, si µs = 0.5. ¿Cuál es el ángulo máximo de elevaci

Sonja [21]

No se creo es 45.7 de nada

7 0

3 years ago

An investigator collects a sample of a radioactive isotope with an activity of 490,000 Bq.48 hours later, the activity is 110,00

Daniel [21]

Answer:

The correct answer is "22.27 hours".

Explanation:

Given that:

Radioactive isotope activity,

= 490,000 Bq

Activity,

= 110,000 Bq

Time,

= 48 hours

As we know,

⇒ $A = A_0 e^{- \lambda t}$

or,

⇒ $\frac{A}{A_0}=e^{-\lambda t}$

By taking "ln", we get

⇒ $ln \frac{A}{A_0}=- \lambda t$

By substituting the values, we get

⇒ $-ln \frac{110000}{490000} = -48 \lambda$

⇒ $-1.4939=-48 \lambda$

$\lambda = 0.031122$

As,

⇒ $\lambda = \frac{ln_2}{\frac{T}{2} }$

then,

⇒ $\frac{ln_2}{T_ \frac{1}{2} } =0.031122$

⇒ $T_\frac{1}{2}=\frac{ln_2}{0.031122}$

$=22.27 \ hours$

3 0

3 years ago