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
<h2>
Explanation:</h2>
<h3>B. Chloroplast</h3>
<em>In </em><em>photosynthesis</em><em> </em><em><u>chloroplast</u></em><em> </em><em>inside</em><em> the</em><em> </em><em>leaf </em><em>contain</em><em> </em><em>chlorophyll</em><em> </em><em>which</em><em> </em><em>captures</em><em> </em><em>light</em><em> </em><em>energy</em><em> </em><em>for</em><em> </em><em>photosynthesis.</em>
<em><u>hope</u></em><em><u> this</u></em><em><u> helps</u></em><em><u> you</u></em>
<em><u>have</u></em><em><u> a</u></em><em><u> good</u></em><em><u> </u></em><em><u>day.</u></em>
A: action force ... to which there is a reaction force