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
Answer: There is not work done at the door because the door did not move.
Explanation: Work is defined as the movement done by a force.
So if you move to apply a force F in an object and you move it a distance D, the work applied on the object is
W = F*D
In this case, the secret agent pushes against the door, so there is a force, but the agent does not move the door, so D = 0, so there is no motion of the door, which implies that there is no work done at the door.
Answer:
Explanation:
The charge alters that space, causing any other charged object that enters the space to be affected by this field. The strength of the electric field is dependent upon how charged the object creating the field is and upon the distance of separation from the charged object.
Answer:
The new length is 50.00885m
Explanation:
linear thermal expansion coefficient Fe 11.8e-6 /K
The new length can be determined using the following equation:
∆L/L = α∆T, where α is linear thermal expansion coefficient
∆L = Lα∆T = 50(11.8e-6)(35-20) = 0.00885 m
New length = ( 50.000 + 0.00885)m =
New length = 50.00885 m