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Harlamova29_29 [7]
3 years ago
9

Jane does 5 joules of work when she closes her bedroom window. She applies a force of 6 newtons to do the job. How far does she

pull the window? Immersive Reader
(1 Point)
30 m
1.2 m
0.83 m
11 m
Physics
1 answer:
insens350 [35]3 years ago
4 0

<u>Statement</u><u>:</u>

Jane does 5 joules of work when she closes her bedroom window. She applies a force of 6 newtons to do the job.

<u>To </u><u>find </u><u>out:</u>

The displacement of the window when she pulled it.

<u>Solution</u><u>:</u>

  • Work done (W) = 5 J
  • Force (F) = 6 N
  • Let the displacement of the window be s.
  • We know, the formula of work done, i.e., W = Fs
  • Putting the values in the above formula, we get
  • 5 J = 6 N × s
  • or, s = (5 ÷ 6) m
  • or, s = 0.83 m

<u>Answer</u><u>:</u>

She pulls the window by 0.83 m.

Hope you could understand.

If you have any query, feel free to ask.

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Answer:

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Explanation:

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As the heat is being release, this value is negative, and same here happens with the work done, because it's in the surroundings.

Therefore the change in the energy would be:

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(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
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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:

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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

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