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pantera1 [17]
3 years ago
13

Write at least 4 to 5 sentences on the different measurements that helped to calculate the speed of light. There are 3 specific

experiments mentioned in your textbook.
Physics
2 answers:
Lina20 [59]3 years ago
7 0

Answer:

Olaus Roemer, Christian Huygens, and Albert Michelson

(Straight out of physics textbook)

Explanation:

The first experiment that was made to measure the speed of light was by Danish astronomer Olaus Roemer. To do this he made precise measurements of Jupiter’ s moon, comparing his observations. Christian Huygens corrected this variation, using the formula, speed of light=d/t. (d) being the distance traveled and (t) being extra time measured, resulting in the speed of light. Albert Michelson was an American physicist who conducted the most famous experiment. He used mirror arrangement to measure the speed of light. The light is reflected back to the eyepiece when the mirror is at rest. These are three experiments conducted that helped to calculate the speed of light.

iren [92.7K]3 years ago
4 0
The first experiment that was made to attempt to measure the speed of light involved detonating gunpowder by Isaac Beeckman. He declared that his experiment was inconclusive. Galileo also tried to measure the speed of light using two lanterns placed across each other. The next experiment involved planets where Ole Romer based his calculation on its movements. The final calculations were derived from different theories by different scientists including Maxwell until it ended up with the exact value for the speed of light. 
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calculate the value of net force acting on given body?5N acting towards right & 15N force acting towards left on the brick.(
lys-0071 [83]

Answer:

The net force acting on the body is 10N directed to the left.

Explanation:

   Magnitude of force to the right = 5N

   Magnitude of force to the left  = 15N

Net force acting on the object and in what direction;

Solution:

It is the vector sum of all forces acting on a body. This net force is the single force that will replace the forces acting on a body;

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   Net force  = Force to the left + Force to the right

 Let us take left to be negative and right to be positive;

    Force to the left  = -15N

    Net force  = -15N + 5N  = -10N

The net force acting on the body is 10N directed to the left.

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3 years ago
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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
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