What collides and creates a movement of heat called conduction?

Softa [21]

The process of soil formation involves the following steps: Accumulation of materials, leaching and losses (because of weather factors particles are leached and eroded away or taken up from the soil by plants), Transformation and illluviation (the chemical weathering<span> of silt, sand, and the formation of clay minerals as well as the change of organic materials into decay resistant organic matter), p</span>odsolisation and translocations (strong acidic solutions breakdown the clay minerals). From the given options soil formation involves both <span>physical and chemical weathering of rocks. Correct answer:D</span>

6 0

3 years ago

It requires 3480 J to move a 675 N object how far?

goblinko [34]

5.16 meters!!!!!!!!!!!!!!!!!!!!

4 0

2 years ago

(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i

murzikaleks [220]

Answer:

a) $t = -\frac{ln(2)}{k}$

b) See the proof below

$A(t) = A_o 2^{-\frac{t}{T}}$

c) $t = 3T \frac{ln(2)}{ln(2)}= 3T$

Explanation:

Part a

For this case we have the following differential equation:

$\frac{dA}{dt}= kA$

With the initial condition $A(0) = A_o$

We can rewrite the differential equation like this:

$\frac{dA}{A} =k dt$

And if we integrate both sides we got:

$ln |A|= kt + c_1$

Where $c_1$ is a constant. If we apply exponential for both sides we got:

$A = e^{kt} e^c = C e^{kt}$

Using the initial condition $A(0) = A_o$ we got:

$A_o = C$

So then our solution for the differential equation is given by:

$A(t) = A_o e^{kt}$

For the half life we know that we need to find the value of t for where we have $A(t) = \frac{1}{2} A_o$ if we use this condition we have:

$\frac{1}{2} A_o = A_o e^{kt}$

$\frac{1}{2} = e^{kt}$

Applying natural log we have this:

$ln (\frac{1}{2}) = kt$

And then the value of t would be:

$t = \frac{ln (1/2)}{k}$

And using the fact that $ln(1/2) = -ln(2)$ we have this:

$t = -\frac{ln(2)}{k}$

Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

$\frac{1}{8} = 2^{-\frac{t}{T}}$

We can apply natural log on both sides and we got:

$ln(\frac{1}{8}) = -\frac{t}{T} ln(2)$

And if we solve for t we got:

$t = T \frac{ln(8)}{ln(2)}$

We can rewrite this expression like this:

$t = T \frac{ln(2^3)}{ln(2)}$

Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

8 0

3 years ago

When water has a temperature greater than 100c?

Ganezh [65]

When water has a temperature greater than 100c, it's boiling

6 0

3 years ago

A truck heading east has an initial velocity of 6 m/s. It accelerates at 2 m/s2 for 12 seconds. What distance does the truck tra

Natali5045456 [20]

-- The truck begins the scenario moving at 6 m/s.

-- After increasing its speed at the rate of 2 m/s² for 12 seconds,
it has gained (2 x 12) = 24 m/s of speed, and it's then moving
at 30 m/s.

-- The truck's average speed during the 12 seconds is (1/2) (6 + 30) = 18 m/s

-- Traveling at an average speed of 18 m/s for 12 seconds,
the truck travels

(18 m/s) x (12 sec) = 216 meters

5 0

3 years ago

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