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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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somewhere between the earth and the moon is a point where the gravitational attraction of the earth is canceled by the gravitati
mote1985 [20]
<span>It's pretty easy problem once you set it up.

Earth------------P--------------Moon

"P" is where the gravitational forces from both bodies are acting equally on a mass m

Let's define a few distances.
Rep = distance from center of earth to P
Rpm = distance from P to center of moon
Rem = distance from center of earth to center of moon

You are correct to use that equation. If the gravitational forces are equal then

GMearth*m/Rep² = Gm*Mmoon/Rpm²

Mearth/Mmoon = Rep² / Rpm²

Since Rep is what you're looking for we can't touch that. We can however rewrite Rpm to be

Rpm = Rem - Rep

Mearth / Mmoon = Rep² / (Rem - Rep)²

Since Mmoon = 1/81 * Mearth
81 = Rep² / (Rem - Rep)²

Everything is done now. The most complicated part now is the algebra, so bear with me as we solve for Rep. I may skip some obvious or too-long-to-type steps.

81*(Rem - Rep)² = Rep²
81*Rep² - 162*Rem*Rep + 81*Rem² = Rep²
80*Rep² - 162*Rem*Rep + 81*Rem² = 0

We use the quadratic formula to solve for Rep:
Rep = (81/80)*Rem ± (9/80)*Rem
Rep = (9/8)*Rem and (9/10)*Rem

Obviously, point P cannot be 9/8 of the way to the moon because it'll be beyond the moon. Therefore, the logical answer would be 9/10 the way to the moon or B.

Edit: The great thing about this idealized 2-body problem, James, is that it is disguised as a problem where you need to know a lot of values but in reality, a lot of them cancel out once you do the math. Funny thing is, I never saw this problem in physics during Freshman year. I saw it orbital mechanics in my junior year in Aerospace Engineering. </span> sylent_reality · 8 years ago
8 0
2 years ago
4. A 505-turn circular-loop coil with a diameter of 15.5 cm is initially aligned so that
Basile [38]

The strength of the magnetic field is 4.8\cdot 10^{-5} T

Explanation:

According to Faraday's Law, the magnitude of the induced emf in the coil is equal to the rate of changeof the flux linkage through the coil:

\epsilon = \frac{N\Delta \Phi}{\Delta t} (1)

where

N = 505 is the number of turns in the coil

\Delta \Phi is the change in magnetic flux through the coil

\Delta t = 2.77 ms = 2.77\cdot 10^{-3} s is the time interval

\epsilon = 0.166 V

The coil is rotated from a position perpendicular to the Earth's magnetic field to a position parallel to it, so the final flux is zero, and the magnitude of the flux change is simply equal to the initial flux:

\Delta \Phi = B A cos \theta

where

B is the strength of the magnetic field

A is the area of the coil

\theta=0^{\circ} is the angle between the normal to the coil and the field

The area of the coil can be written as

A=\pi r^2

where

r=\frac{15.5 cm}{2}=7.75 cm = 7.75\cdot 10^{-2} m is its radius

Substituting everything into eq.(1) and solving for B, we find:

\epsilon= \frac{NB\pi r^2 cos \theta}{\Delta t}\\B=\frac{\epsilon \Delta t}{\pi r^2 cos \theta}=\frac{(0.166)(2.77\cdot 10^{-3})}{(505)\pi (7.75\cdot 10^{-2})^2(cos 0^{\circ})}=4.8\cdot 10^{-5} T

Learn more about magnetic fields:

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

8 0
2 years ago
Read 2 more answers
Two sinusoidal waves travel along the same string. They have the same wavelength and frequency. Their amplitudes are ym1 = 2.5 m
Nimfa-mama [501]

Answer:

0.5 m

Explanation:

Givens:

ym1 = 2.5 mm

ym2 = 4.5 mm

Ф_1=π / 4

Ф_2=π / 2

We have 2 ways to solve this problem. The first one given that the 2 waves have the frequency then we know that the resultant wave amplitude is

Ym = (ym1 + ym2)cos(Ф_2/2)

By substitution we have  

Ym= (0.025 + 0.045)cos(π/4) = 0.496 m

The second one is it treat them as Phasors where the phase between them is Ф_2=π / 2 Therefore  

Ym^2=(ym1^2+ym2^2)

So we have Ym=√0.025^2+0.045^2

                         = 0.5 m

7 0
2 years ago
Suppose the truck that’s transporting the box In Example 6.10 (p. 150) is driving at a constant speed and then brakes and slows
Scorpion4ik [409]

Answer:

Friction acts in the opposite direction to the motion of the truck and box.

Explanation:

Let's first review the problem.

A moving truck applies the brakes, and a box on it does not slip.

Now when the truck is applying brakes, only it itself is being slowed down. Since the box is slowing down with the truck, we can conclude that it is friction that slows it down.

The box in the question tries to maintains its velocity forward when the brakes are applied. We can think of this as the box exerting a positive force relative to the truck when the brakes are applied. When we imagine this, we can also figure out where the static friction will act to stop this positive force. Friction will act in the negative direction. Or in other words, friction will act in the opposite direction to the motion of the truck and box. This explains why the box slows down with the truck, as friction acts to stop its motion.

5 0
1 year ago
Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat
Alja [10]

Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

Note: the mass doesn't change, it is only the velocity that change.

Also,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Let assume that if is constant acceleration

Then, m and a is constant,

Then,

K.E is directly proportional to d

So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

K.E_1 = E_k

d_1 = d

d_2 = d/3

K.E_2 = K.E_1 / d_1 × d_2

K.E_2 = E_k × ⅓d / d

Then,

K.E_2 = ⅓E_k

So, the new kinetic energy is one third of the E_k

7 0
2 years ago
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