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irga5000 [103]
4 years ago
11

myron is almost late for class and he is running quickly to arrive before the professor begins lecturing as he listen to the pro

fessor lecture myron notices that his heart started beating rapidly and his respiration rate is quite high myron conclude that he must be interested in the lecture abd professor because he would not be so aroused otherwise the theory would best explain myron's conclusion
Physics
1 answer:
amid [387]4 years ago
6 0
There is no theory here.
Myron has offered one hypothesis to explain his observations.
There are other possible hypotheses.
They include:

-- An infected mosquito might have bitten him while he slept,
and the results of the infection might be starting to show up.

-- He might have eaten something for dinner last night
that was slightly spoiled.

-- He might have imbibed too much beer for his own good
at the fraternity party last night.

-- There may be too much Carbon Dioxide in the classroom air.

-- His body may be reacting to the physical stress of running to class.

So far, Myron only has a hypothesis.
He's in no position to come to any "conclusion" until he tests
his hypothesis, and shows that the same results follow the same
conditions MOST of the time.  His hypothesis may be difficult to 
test, but until he does that, he doesn't have a theory.

My personal opinion is that while his hypothesis may also be correct,
the most likely source of his observation is the recent physical stress 
of running to class.  It's important to understand that I'm in no position
to try and convince anyone of this conclusion.  My opinion is simply
another hypothesis.  It carries no weight until it's tested.
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Astronauts on a distant planet set up a simple pendulum of length 1.2 m. The pendulum executes simple harmonic motion and makes
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An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f
IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}

Similarly, The equation of mption:

x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})

replacing our assumed values:

where v_=0 \ and \ g= 9.81

x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})

x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

1000= 0.981t-0.981(1-e^{-(10)t}) \ m

By diregarding e^{-(10)t} \ m

1000= 0.981t-0.981

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0
3 years ago
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