If they both have the same kinetic energy then they both have the same kinetic energy. otherwise it would be the bowling ball due to factors like weight
The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.
Answer:
D. At x=0, it's acceleration is at a maximum
Explanation:
As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.
Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.
From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.
The same can be said as the box travels backward from point A to -A
Answer:
Hello your question is poorly written below is the complete question
Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?
answer :
a) 231.48 days
b) n = 3.125 * 10^15
Explanation:
Battery moved 10,000 coulombs
current rate = 0.5 mA
<u>A) Determine how long the clock run on the battery. use the relation below</u>
q = i * t ----- ( 1 )
q = charge , i = current , t = time
10000 = 0.5 * 10^-3 * t
hence t = 2 * 10^7 secs
hence the time = 231.48 days
<u>B) Determine how many electrons per second flowed </u>
q = n*e ------ ( 2 )
n = number of electrons
e = 1.6 * 10^-19
q = 0.5 * 10^-3 coulomb ( charge flowing per electron )
back to equation 2
n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )
hence : n = 3.125 * 10^15
The net force is 12 N to the left.
In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:
The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:
The wavelength of the 2nd harmonic is:
The wavelength of the 4th harmonic is:
It is not possible to find any integer n such that 
, therefore the correct options are A, B and D.
