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2 years ago

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The stone, which weighs 400 g, is thrown upwards at a speed of 20 m / s. Climbed to a height of 12 m. Determine: what is equal t

o the work of the resistance force; equal to the mean resistance force?

1 answer:

maxonik [38]2 years ago

3 0

Given that,

Mass of the stone, m = 400 g = 0.4 kg

Initial speed, u = 20 m/s

It is climbed to a height of 12 m.

To find,

The work done by the resistance force.

Solution,

Let v is the final speed. It can be calculated by using the conservation of energy.

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 12} \\\\v=15.33\ m/s$

Work done is equal to the change in kinetic energy. It can be given as follows :

$W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 0.4\times (15.33^2-20^2)\\\\=-32.99\ J$

So, the required work done is 32.99 J.

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$m_1v_1+m_2v_2=(m_1+m_2)v'$

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Answer:

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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:

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

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$ln |A|= kt + c_1$

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:

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

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

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

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

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:

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

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