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Maurinko [17]
3 years ago
6

If you were traveling away from Earth at speed 0.5c, would you notice a change in your heartbeat? Would your mass, height, or wa

istline change? What would observers on Earth using telescopes say about you?
Physics
1 answer:
FrozenT [24]3 years ago
8 0

Answer:

If you were traveling away from earth at speed 0.5c, you wouldn't notice any change in your heartbeat, you won't notice your mass, height and waistline change. This is because you are on the same frame of reference as the ship in spacetime and any measurement done from the ship will give normal readings from an observer on the ship.

For an observer on earth, your heartbeat will be seen to slowdown (because your time on the ship will be perceived to slow down to an

observer on earth). Also, your mass will be seen to increase, you height will also be seen to increase, and your waistline will be seen to decrease when viewed from earth.

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This problem must be solved using a sketch. I attached an illustration of the problem.
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Law of reflection tells us that light ray reflects off the surface at the same angle in which it falls on it( i attached another illustration of this).
Now we can write tangens equations:
tan(\theta)=\frac{h-0.8}{1}\\
tan(\theta)=\frac{1.8-h}{2.8}\\
\frac{h-0.8}{1}=\frac{1.8-h}{2.8}
We solve for h:
\frac{h-0.8}{1}=\frac{1.8-h}{2.8}\\
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3 years ago
Newton’s empirical law of cooling/warming of an object is given by ( ), T Tm k dt dT = − where k is a constant of proportionalit
pychu [463]

Answer:

The time for the cake to cool off to room temperature is

approximately 30 minutes.

Let T_{0} = 70^{0}F be the temperature and T that of the body

Explanation:

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\frac{DT}{dt} = <em>k</em>(T − 70), T(0) = 300

Solving the equation, we get

\frac{DT}{t-70} = <em>kdt</em>

In [T-70]= <em>kt </em>+C_{1}

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Finding he value for C_{2} using the initial value of T (0)= 300, therefore we get:

300=70+C_{2}

C_{2} = 230 therefore

T= 70+ 230 e^{kt}

Finding the value for <em>k </em>using T (3)  = 200, therefore we get

T (3) = 200

e^{3k} = \frac{13}{23}

<em>K </em>= \frac{1}{3} in \frac{13}{23}

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T(t) = 70+230e^{-0.19018}

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3 years ago
A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts help.
GrogVix [38]

Answer:

H = 532 m

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this total distance must be equal to twice the height of the cliff

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