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kap26 [50]
3 years ago
15

Why is genetics considered a theory and not a hypothesis?

Physics
2 answers:
lukranit [14]3 years ago
6 0
Theory is a guess that consists of something a group of people believe to be true a hypothesis is an educated guess

Paha777 [63]3 years ago
3 0
I dont know i asked that question?!?!?!?!?!?
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DESCRIBE THE FORMATION OF THE SOLAR SYSTEM ACCORDING TO THE NEBULAR THEORY
marishachu [46]

When a cloud of gas and dust in space was disturbed, maybe by the explosion of a nearby star.This explosion made waves in space which squeezed the cloud of gas & dust.

8 0
3 years ago
Insert
nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

A)

The activity of a radioactive sample is the number of decays per second in the sample.

The activity of a sample is therefore directly proportional to the number of nuclei in the sample:

A\propto N

where A is the activity and N the number of nuclei.

As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:

A\propto m

where m is the mass of the sample.

In this problem:

- When the mass is m_1 = 1 g, the activity is A_1=16 count/min

- When the mass is m_2=20 g, the activity is A_2

So we can find A2 by using the rule of three:

\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min

B)

The equation describing the activity of a radioactive sample as a function of time is:

A(t)= A_0 e^{-\lambda t} (1)

where

A_0 is the initial activity at time t = 0

t is the time

\lambda is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

\lambda = \frac{ln2}{t_{1/2}}

where t_{1/2} is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.

In this problem, carbon-14 has half-life of

t_{1/2}=5700 y

So its decay constant is

\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

A_0 = 320 count/min (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min

C)

We know that a sample of living wood has an activity of

A=16 count/min per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

A_0 = A\cdot 5 = (16)(5)=80 count/min

Moreover, here we have a sample of 5 g, with current activity of A=20 count/min: it means that its activity per gram of mass is

A'=\frac{20}{5}=4 count/min

We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:

- After 1 half-life, the activity drops from 16 count/min to 8 count/min

- After 2 half-lives, the activity dropd to 4 count/min

So the age of the wood is equal to 2 half-lives, which is:

t=2t_{1/2}=2(5700)=11,400 y

3 0
4 years ago
What simple machine can best be described as "a simple machine that uses an inclined plane wrapped around a rod"?
Alekssandra [29.7K]
A screw I think I’m sorry if you get it wrong
7 0
3 years ago
Read 2 more answers
A particle moves along the x-axis with velocity given by v(t)=3t2+6t for time tâ¥0. If the particle at position x=2 at time t=0,
sergey [27]

Answer:

b) 6

Explanation:

Given

v(t)=3t²+6t

X(0) = 2

X(1) = ?

Knowing that

v(t)=3t²+6t = dX/dt

⇒ ∫dX = ∫(3t²+6t)dt

⇒ X - X₀ = t³ + 3t²

⇒ X(t) = X₀ + t³ + 3t²

If  X(0) = 2

⇒  X(0) = X₀ + (0)³ + 3(0)² = 2

⇒  X₀ = 2

then we have

X(t) = t³ + 3t² + 2

when

t = 1

X(1) = (1)³ + 3(1)² + 2

X(1) = 6

8 0
4 years ago
Read 2 more answers
An object that is dropped from a height H falls with a constant acceleration of g. The final
GuDViN [60]

The height need to change by 4 to double the final velocity.

<h3>Final velocity of the object</h3>

The final velocity of an object during a free fall is related to maximum height of fall as given the equation below.

v = √2gh

v² = 2gh

v²/h = 2g

v₁²/h₁ = v₂²/h₂

when v₂ = 2v₁, change in height is calculated as;

h₂ = h₁v₂²/v₁²

h₂ = (h₁ (2v₁)²) / (v₁)²

h₂ = 4h₁v₁² / v₁²

h₂ = 4h₁

Thus, the height need to change by 4 to double the final velocity.

Learn more about final velocity here: brainly.com/question/25905661

#SPJ1

4 0
2 years ago
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