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
(1)
Cheetah speed: 
Its position at time t is given by
(1)
Gazelle speed: 
the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by
(2)
The cheetah reaches the gazelle when 
. Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:
(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.
Cheetah: 
Gazelle: 
So, the gazelle should be ahead of the cheetah of at least
Answer:
a
Explanation:
as the copper wire is very dangerous so now if these two thing happens then it would easily help the current flows through it so it might be a little bit easy for the current to flow through it
Answer:
C
Explanation:
Gravity is the main reason that make our planets to pull each other
Answer:
The answer is True
Explanation:
Statistical Multiplexing is considered an example of communication link sharing which makes it comparable to DBA (Dynamic Bandwidth Allocation). Here, communication channels are broken down into data streams to optimize the communication process.
In Statistical Time-division Multiplexing, time slots are allocated to data streams for communication optimization. This method makes sure that no time slot or bandwidth is wasted.
Hence, the sum of combined circuits must not be equal to the capacity of the circuit to work effectively.
