Answer:
The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.
Explanation:
Given that,
Mass = 2 kg
Radius = 0.5 m
Angular speed = 3 rad/s
Force = 10 N
(I). We need to calculate the rotational kinetic energy
Using formula of kinetic energy
(II). We need to calculate the instantaneous change rate of the kinetic energy
Using formula of kinetic energy
On differentiating
....(I)
Using newton's second law
Put the value of a in equation (I)
Hence, The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.
Answer:
<h2>2500 N</h2>
Explanation:
The weight of the object can be found by using the formula
w is the workdone
d is the distance
From the question we have
We have the final answer as
<h3>2500 N</h3>
Hope this helps you
He did sadly can’t believe it T-T
Answer:
Explanation:
According to Kepler's second law: a line segment joining an object and the Sun sweeps out equal areas during equal intervals of time. This law is equivalent to the constancy of the angular momentum, that is to say:
Here, is the comet's mass, is its distance from the Sun, is its speed and the angle between and
The aphelion and the perihelion are the only two points in the orbit where the radius and velocity are perpendicular. Therefore, at these 2 points, the angular momentum module can be calculated as a simple product:
Rewriting for :
Answer:
102900 Joules
Explanation:
Assuming the kinetic energy was zero at the moment of release, you can make the following argument to solve the problem:
The potential energy at full height was mgh. We are told that after 70% of the distance, i.e., mg(0.3h) = 44.1kJ. Since potential energy is linear in altitude h, we get get the full potential energy to be 44.1kJ/0.3. The difference between full potential energy and the one after 70% of the way must equal the gained kinetic energy (neglecting stuff like heat due to friction). So,
44.1kJ/0.3 - 44.1kJ = 0.7*44.1kJ/0.3 = 102.9kJ = Ekinetic
The kinetic energy after 70% of the falling distance was 102.9 kJ.