The answer is <span> A. Yes, the wave will transfer the energy of Amy's motion to the other end of the rope.</span>
The kinetic energy at the bottom of the swing is also 918 J.
Assume the origin of the coordinate system to be at the lowest point of the pendulum's swing. A pendulum, when raised to the highest point has potential energy since it is raised to a height h above the origin. At the highest point, the pendulum's velocity becomes zero, hence it has no kinetic energy. Its energy at the highest point is wholly potential.
When the pendulum swings down from its highest position, it gains velocity. Hence a part of its potential energy begins to convert itself into kinetic energy. If no dissipative forces such as air resistance exist, then, the law of conservation of energy can be applied to the swing.
Under the action of conservative forces, the total mechanical energy of a system remains constant.This means that the sum of the potential and kinetic energies of a body remains constant.
When the pendulum reaches the lowest point of its swing, it is at the origin of the chosen coordinate system. Its vertical displacement from the origin is zero, hence its potential energy with respect to the origin is zero. Therefore the entire potential energy of 918 J should have been converted into kinetic energy, according to the law of conservation of energy.
Thus, the kinetic energy of the pendulum at the lowest point of its swing is equal to the potential energy it had at its highest point, which is equal to <u>918 J.</u>
Answer:
Explanation:
Let the race be of a fixed distance x
![Average Speed = \frac{Total Distance}{Total Time}](https://tex.z-dn.net/?f=Average%20Speed%20%3D%20%5Cfrac%7BTotal%20Distance%7D%7BTotal%20Time%7D)
Troy's Average speed = 3 miles/hr = x / 0.2 hr
x = 0.6 miles
Abed's Average speed = 0.6 / 0.125 = 4.8 miles/hr
Lamina occupies x² + y² = 14y. Outside circle is x² + y² = 49
To find the mass of lamina, integrate given density function over the region
m = ∫∫D P(x, y) dA
Subtitute x = r cosФ and y = r sinФ in x² + y² = 14y
and x² + y² = 49
x² + y² = 14y
(r cosФ)² + ( rsinФ)² = 14(rsinФ)
r² = 14r sin Ф
x² +y² = 49
r² = 49
r = 7
Cntre mass (-x. -y)
-x= i/m ∫∫D xp(x,y) dA = 1/m∫∫ (r cosФ) p( r, Ф)r
dr dФ
-y = 1/m∫∫D yp(x, y) dA = 1/m ∫∫D (r sinФ) p(r, Ф) r drФ
where m = ∫∫D p(x, y) dA
<u>Astronauts are not weightless during either launch or return to Earth.</u>
<u></u>
<h3>
Brief explanation</h3>
Astronauts become weightless when they stop accelerating. Basically that means when the engines cut out and they begin to coast in orbit. They will remain “weightless” for as long as they are in orbit. To get out of orbit, they have to decelerate (i.e. Accelerate in the opposite direction) and so they begin to feel a force that feels very much like gravity as they are falling back to Earth.
One of the cool things is that you can't tell the difference between gravity and acceleration. They're the same thing as far as your body is concerned. Einstein used a variety of somewhat related thought experiments while he has working out the details of the special theory of relativity.
Hence, with this explanation , we can conclude that astronauts are not weightless during either launch or return to Earth.
Learn more about astronauts being weightless
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