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2 years ago

Which of the following does NOT support the geologic evidence used to date the earth?

Physics

1 answer:

Zanzabum2 years ago

4 0

Answer:

comparing the anatomy of organisms

Explanation:

Basically, Rocks are the solid evidence used to date the earth.

Due to the changing in the genetic make up of organisms,the anatomy of organisms is not a reliable evidence to date the earth.

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Vector ~A has a magnitude of 29 units and points in the positive y-direction. When vector ~B is added to ~A, the resultant vecto

timurjin [86]

Answer:

The magnitude of vector B is 43 units and it points in the negative y-direction.

Explanation:

Resultant of vectors = vector sum of all the vectors

Vector A = 29j

Vector B = ?

Resultant of vector A and B = R = -14j

R = A + B

-14j = 29j + B

B = -14j - 29j = - 43j

Hence, the magnitude of vector B is 43 units and it points in the negative y-direction.

4 0

2 years ago

A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function

Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$$v(t)=\sin (\omega t) * \cos ^2(\omega t)$

To get the position equation we just need to integrate the above equation:

$$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t$

$$\mathrm{u}=\cos (\omega \mathrm{t})$

Then:

$$d u=-\sin (\omega t) d t$

$\Rightarrow d t=-d u / \sin (\omega t)$

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$$f(t)=\frac{\cos ^3(\omega t)}{3}+C$

To find the value of C, we use the fact that f(0) = 0.

$$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0$

C = -1 / 3

Then the position function is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

4 0

11 months ago

Ablock of mass m2 on arough horinzontal surfaceis connected to aball of mass m1 by alight weight cord over alight weight frictio

aliya0001 [1]

Dndncbsnsndkdkksdkdndnsndnkckd

5 0

2 years ago

Calculate the speed of a car that travels a distance of 2.4km in 2 minutes

hammer [34]

Answer:

72 kilometres per hour

Explanation:

The formula for calculating speed is distance/time.

So to work this out you would convert 2 minutes into hours. You would divide 2 by 60 to convert it into hours. This is because the standard unit for speed with kilometres is kilometres per hour. Then you would divide 2.4 by that.

1)Divide 2 by 60:

$2/60=0.0333333$

2) Divide 2.4 by 0.0333333.

$2.4/0.0333333=72.00007$

3) Round it.

$72km/h$

3 0

2 years ago

What is the volume V of a sample of 4.00 mol of copper? The atomic mass of copper (Cu) is 63.5 g/mol, and the density of copper

rusak2 [61]

Answer : The volume of a sample of 4.00 mol of copper is $28.5cm^3$

Explanation :

First we have to calculate the mass of copper.

$\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}$

$\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g$

Now we have to calculate the volume of copper.

Formula used :

$Density=\frac{Mass}{Volume}$

Now put all the given values in this formula, we get:

$8.92\times 10^3kg/m^3=\frac{254g}{Volume}$

$Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3$

Conversion used :

$1kg/m^3=1g/L\\\\1L=10^3cm^3$

Therefore, the volume of a sample of 4.00 mol of copper is $28.5cm^3$

7 0

2 years ago

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