During one trial, the acceleration is 2m/s^2 to the right. What calculation will give the tensions in actin filaments during thi
1 answer:
Answer: hello your question is poorly written attached below is the complete question
answer :
TA = 1.6*10^-24 * 60 * 2, TB = 1.6*10^-24 * ( 60 + 30 ) * 2 -- ( option 1 )
Explanation:
a = 2m/s^2
Ta = m₁ a = 60 * 1.6 * 10^-24 * 2 ц
Tb - Ta = m₂ a
∴ Tb = m₂ a + Ta
= ( 30 * 1.6 * 10^-24 * 2 ) + ( 60 * 1.6 * 10^-24 * 2 )
= ( 30 + 60 ) * 1.6 * 10^-24 * 2 ц
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I would have to see the graph.. but by looking at one one online, they are between points D and E.
<h2>Answers:</h2>
:
>>>><u>This case</u></h2>
(a) The kinetic energy of a body is that energy it possesses due to its movement and is defined as:
(1)
Where is the mass of the body and
its velocity.
In this specific case of the satellite, its kinetic energy taking into account its mass
is:
(2)
On the other hand, the velocity of a satellite describing a circular orbit is constant and defined by the following expression:
(3)
Where is the gravity constant,
the mass of the Earth and
the radius of the orbit <u>(measured from the center of the Earth to the satellite).
</u>
Now, if we substitute the value of from equation (3) on equation (2), we will have the final expression of the kinetic energy of this satellite:
(4)
Finally:
(5) >>>>This is the kinetic energy of the satellite
(b) According to Kepler’s 2nd Law applied in the case of a circular orbit, its Period is defined as:
(6)
Where is a constant and is equal to
. So, this equation in these terms is written as:
(7)
As we can see, <u>the Period of the orbit does not depend on the mass of the satellite </u>, it depends on the mass of the greater body (the Earth in this case)
, the radius of the orbit and the gravity constant.
(c) The gravitational force described by the law of gravity is a central force and therefore is <u>a conservative force</u>. This means:
1. The work performed by a gravitational force to move a body from a position A to a position B <u>only depends on these positions and not on the path followed to get from A to B. </u>
2. When the path that the body follows between A and B is a c<u>losed path or cycle</u> (as this case with a <u>circular orbit</u>), <u>the gravitational work is null or zero</u>.
<h2>This is because the gravity force that maintains an object in circular motion is a centripetal force, that is, <u>it always acts perpendicular to the movement</u>. </h2>
Then, in the case of the satellite orbiting the Earth in a circular orbit, its movement will always be perpendicular to the gravity force that attracts it to the planet, at each point of its path.
(d) The total Mechanical Energy of a body is the sum of its Kinetic Energy
and its Potential Energy
:
(8)
But in this specific case of the circular orbit, its kinetic energy will be expresses as calculated in the first answer (equation 5):
(5)
And its potential energy due to the Earth gravitational field as:
(9)
This energy is negative by definition.
So, the total mechanical energy of the orbit, also called the Orbital Energy is:
(10)
Solving equation (10) we finally have the Orbital Energy:
(11)
At this point, it is necessary to clarify that a satellite (or any other celestial body) orbiting another massive body, can describe one of these types of orbits depending on its Orbital Total Mechanical Energy :
-When :
We are talking about an <u>open orbit</u> in which the satellite escapes from the attraction of the planet's gravitational field. The shape of its trajectory is a parabola, fulfilling the following condition:
Such is the case of some comets in the solar system.
-When :
We are also talking about <u>open orbits</u>, which are hyperbolic, being
<h2>-When
We are talking about <u>closed orbits</u>, that is, the satellite will always be "linked" to the gravitational field of the planet and will describe an orbit that periodically repeats with a shape determined by the relationship between its kinetic and potential energy, as follows:
-Elliptical orbit: Although is constant,
and
are changing along the trajectory
.
-Circular orbit: When at all times both the kinetic energy and the potential
remain constant, resulting in a total mechanical energy
as the one obtained in this exercise. This means that the speed is constant too and <u>is the explanation of why this Energy has a negative sign.
</u>