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beks73 [17]
2 years ago
10

Suppose the battery in a clock wears out after moving 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?
Physics
1 answer:
Ksivusya [100]2 years ago
8 0

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

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<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2><h3>DATA:</h3>

Height of Aeroplane = 500 meters

Speed of Aeroplane = 200 m/s

Acceleration due to gravity = g = 10m/s^2

time of package to reach the ground = t = ?

Horizontal distance = d = ?

<h2>_____________________________________</h2><h3>SOLUTION:</h3>

When the bomb is dropped, it will have an initial horizontal velocity which is equal to the speed of the plane as the plane is just moving in horizontal direction. So the bomb fall and will travel forward.

Initial horizontal velocity is given by,

                                            V_{ix} = 200 ms{-1}

Initial vertical velocity is given by,

                                            V_{iy} = 0m/s

By the second equation of motion under gravity,

                                           H = V_{iy}t + \frac{1}{2}gt^2\\\\500 = (0)t + \frac{1}{2}x10xt^2 \\\\500 = 0 + 5t^2\\\\t^2 = \frac{500}{5}\\\\t^2 = 100\\\\t = 10 seconds

<h2>_____________________________________</h2>

<u>For horizontal distance(d)</u>:

Horizontal distance has no acceleration thus d is given by

                                          d = vt

                                          d = 200x10

                                          d = 2000 meters

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

                 

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