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3 years ago

(this is somehow part of my science unit, dont ask why)

Physics

1 answer:

nirvana33 [79]3 years ago

6 0

Answer:

Height of coffee mug = 5 cm

Explanation:

Changing all the data given in to unit cm

50 mm = 5 cm

20 cm = 20 cm

8 m = 800 cm

0.1 km = 100 m = 10000 cm

Since we need to find the height of coffee mug, it's height will not be more than 10 cm.

So 5 cm is the correct choice.

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Pls help mee

NISA [10]

Answer:

the Arrow X shows the direction of amplitude

Explanation:

As the amplitude is the maximum displacement of a wave from the mean position

8 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

What type of transfer is affected by color

Mama L [17]

Heat transfer is affected by color.

7 0

3 years ago

Describe the motion of an object that has an acceleration of 0m/s2

solong [7]

Zero acceleration means the object's velocity is not changing.
So the object is moving in a straight line, at a constant speed
that could be anything (including zero) as long as it's constant.

6 0

3 years ago

A 0.5 kg yellow bouncy ball rolls to the right at a speed of 8 m/s while a 0.25 kg blue bouncy ball rolls

inessss [21]

Explanation:

Let yellow ball be m1 = 0.5kg with u1 = 8 m/s and blue ball be m2 = 0.25 kg with u2 = - 4 m /s respectively.

After collision, blue ball travels 12 m/s.

<u>Using conservation of Linear Momentum</u> :

m1u1 + m2u2 = m1v1 + m2v2

0.5* 8 + 0.25 * - 4 = 0.5 * v1 + 0.25 * 12

v1 = 0 m/sec i.e. <u>Yellow ball comes to rest</u>.

3 0

3 years ago