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3 years ago

11

Zona onde não incide luz visível devido à interposição de um objeto opaco.

1 answer:

m_a_m_a [10]3 years ago

8 0

Answer:

can you please translate to english?

Explanation:

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Is diverse families positive or negative

nataly862011 [7]

Think its Positive
hope this helpes

6 0

3 years ago

The world's most powerful laser is the LFEX laser in Japan. It can produce a 2 petawatt(2×10^15W) laser pulse that last for 1 ps

kogti [31]

Answer:

$2.82942\times 10^{24}\ W\m^2$

Explanation:

d = Diameter of spot = 30 μm

r = Radius of spot = $\frac{d}{2}=\frac{30}{2}=15\ mu m$

P = Power of the laser = $2\times 10^{15}\ W$

A = Area = $\pi r^2$

Intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{2\times 10^{15}}{\pi\times (15\times 10^{-6})^2}\\\Rightarrow I=2.82942\times 10^{24}\ W/m^2$

The light intensity within this spot is $2.82942\times 10^{24}\ W/m^2$

8 0

3 years ago

Which of the following objects exerts a gravitational force?

kompoz [17]

the earth exerts a gravitational force

4 0

2 years ago

Read 2 more answers

(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

Imagine that you are jumping up and down on a trampoline.

lions [1.4K]

Answer:

<em />

<em>a. When you are pressing downwards</em><u><em> the energy stored is potential elastic energy due to the elasticity of the stretchy skin.</em></u><em> </em>

<em> </em>

<em> </em>

<em> </em>

<em>b. When you are high up in the air </em><u><em>the energy stored is potential energy due to your weight and the distance to the ground.</em></u>

<em />

<em>Hope this Helps!</em>

<em />

<em>If you have Any questions Please Ask.</em>

5 0

2 years ago