<em>Kinetic Energy</em>
=><em><u>It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.</u></em>
<em>Potential</em><em> </em><em>Energy</em><em> </em>
<u><em>=</em><em>></em><em>potential energy is the energy held by </em></u><em><u>an</u></em>
<em><u> object because of its position relative to </u></em><em><u>other</u></em>
<em><u> objects, stresses within itself, its </u></em><em><u>electric</u></em>
<em><u> charge, or other factors.</u></em>
<h2>Difference:</h2>
=>Potential energy is a <u>stored</u> energy on the other hand kinetic energy is the energy of an object or a system's particle in <em><u>Motion</u></em>.
Answer:


Explanation:
Usando la ley de Hook tenemos:

Solving it for k we have:



Usando la misma ecuación y sabiendo k tenemos:



Espero esto te ayude!
Answer: 4.98 m/s
Explanation:
You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.
PEi = 0 in this case
KEi = ½mVi² = PEf+KEf = mghf + ½mVf²
½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²
½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26
Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s
Answer:
option C
Explanation:
The correct answer is option C
There is no external force acting in the system hence the momentum will be conserved.
As the milk is leaking out of the tank mass of the tanker is decreasing.
When the mass of the container will decrease to conservation the momentum speed of the container will have to be increased.
So, the car carrying milk will speed up.
The frequency of the human ear canal is 2.92 kHz.
Explanation:
As the ear canal is like a tube with open at one end, the wavelength of sound passing through this tube will propagate 4 times its length of the tube. So wavelength of the sound wave will be equal to four times the length of the tube. Then the frequency can be easily determined by finding the ratio of velocity of sound to wavelength. As the velocity of sound is given as 339 m/s, then the wavelength of the sound wave propagating through the ear canal is
Wavelength=4*Length of the ear canal
As length of the ear canal is given as 2.9 cm, it should be converted into meter as follows:

Then the frequency is determined as
f=c/λ=339/0.116=2922 Hz=2.92 kHz.
So, the frequency of the human ear canal is 2.92 kHz.