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vredina [299]
3 years ago
11

Anyone wanna join whereby? https://whereby.com/dancing-goddess

Physics
1 answer:
AleksandrR [38]3 years ago
6 0
Sure let me download it rn
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Setting goals that match your lifestyle helps improve chances that they will be reached.
Tasya [4]

Answer:

TRUE!!!!!!

Explanation:

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The graph to the right shows the change in Canada‘s harvest of Atlantic cod from 1950-2004 what year shows the clearest evidence
Ann [662]

The correct answer is C. 1995

Explanation:

The graph shows the changes in the harvest of Atlantic cod. In general, this graph illustrates how the peak occurred in the 1980s but then there was a sudden and sharp decline in 1995. Indeed, 1995 is the year with the lowest number of harvested cod as in this year there were approximately least than 10 thousand metric tonnes of cods. Also, this year shows the collapse of fishing stocks or that the population of this fish collapsed, which made it impossible to harvest as many fish as in previous years. According to this, the year that shows the collapse of fishing stocks is 1995.

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3 years ago
Define chemical and physical changes. (
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Chemical change is two things get chemical reaction and produce a new product.

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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
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
A. What is an electric field line?
Lorico [155]

Answer:

Please see below as the answers are self-explanatory

Explanation:

a)

  • A electric field line is an imaginary line, which has the property that the electric field vector is tangent to it at any point. It starts from positive charges (since the electric field by convention it has the direction of the trajectory that would take a positive test charge, so it always goes away from positive charges) and ends in negative charges.

b)

  • Since the potential difference between two points represents the work per unit charge needed for a charge to move between these points, a equipotential surface is the one over which it is not needed to do work to move a charge from any point on the surface to any other point, which means that all points are at the same potential.

c)

  • Equipotential surfaces are not necessarily physical surfaces, they can be defined in vaccum for instance.
  • As an example, any spherical surface concentric with a point charge, is an equipotential surface, and  it can be a real surface or a fictitious one.
7 0
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