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irga5000 [103]
2 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]2 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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Submarine a travels horizontally at 11.0 m/s through ocean water. it emits a sonar signal of frequency f 5 5.27 3 103 hz in the
xeze [42]
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is 
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
6 0
2 years ago
The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the
Diano4ka-milaya [45]

Answer:

<em>The object could fall from six times the original height and still be safe</em>

Explanation:

<u>Free Falling</u>

When an object is released from rest in free air (no friction), the motion is completely dependant on the acceleration of gravity g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy of

U=m.g.h

When the object strikes the ground, all the mechanical energy (only potential energy) becomes into kinetic energy

\displaystyle K=\frac{1}{2}m.v^2

Where v is the speed just before hitting the ground

If we know the speed v is safe for the integrity of the object, then we can know the height it was dropped from

\displaystyle m.g.h=\frac{1}{2}m.v^2

Solving for h

\displaystyle h=\frac{m.v^2}{2mg}=\frac{v^2}{2g}

If the drop had occurred in the Moon, then

\displaystyle h_M=\frac{v_M^2}{2g_M}

Where hM, vM and gM are the corresponding parameters on the Moon. We know v is the safe hitting speed and the gravitational acceleration on the Moon is g_M=1/6 g

\displaystyle h_M=\frac{v^2}{2\frac{1}{6}g}

\displaystyle h_M=6\frac{v^2}{2g}=6h

This means the object could fall from six times the original height and still be safe

6 0
2 years ago
The lightest duty and most widely used non flexible metal conduit is
Alenkinab [10]
Aluminium, or heavier version copper.
6 0
2 years ago
Two blocks, 1 and 2, are connected by a rope R1 of negligible mass. A second rope R2, also of negligible mass, is tied to block
alekssr [168]

Answer:

Explanation:

Given

Two block are connected by rope R_1

R_2 rope is attached to block 2

suppose F_2 is a force applied to Rope R_2

Applied force F_2=Tension in Rope 2

F_2=(m_1+m_2)a---1

where a=acceleration of system

Tension in rope R_1 is denoted by F_1

F_1=m_1a---2

divide 1 and 2 we get

\frac{F_2}{F_1}=\frac{(m_1+m_2)a}{m_1a}

also m_1=2.11\cdot m_2

\frac{F_2}{F_1}=\frac{2.11m_2+m_2}{2.11m_2}

\frac{F_2}{F_1}=\frac{3.11}{2.11}

\frac{F_1}{F_2}=\frac{2.11}{3.11}

               

3 0
2 years ago
If the force of gravity between a book of mass 0.50 kg and a calculator of 0.100 kg is 1.5 × 10-10 N, how far apart are they?  (
valkas [14]
The gravitational force between two masses m₁ and m₂ is
F=G \frac{m_{1} m_{2}}{d^{2}}
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d =  distance between the masses.

Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg

Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
     = 0.0222
d = 0.1492 m = 149.2 mm

Answer: 149.2 mm
8 0
2 years ago
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