<h2>
Answer:</h2>
(a) 3.96 x 10⁵C
(b) 4.752 x 10⁶ J
<h2>
Explanation:</h2>
(a) The given charge (Q) is 110 A·h (ampere hour)
Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;
=> Q = 110A·h
=> Q = 110 x 1A x 1h [1 hour = 3600 seconds]
=> Q = 110 x A x 3600s
=> Q = 396000A·s
=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C
Therefore, the number of coulombs of charge is 3.96 x 10⁵C
(b) The energy (E) involved in the process is given by;
E = Q x V -----------------(i)
Where;
Q = magnitude of the charge = 3.96 x 10⁵C
V = electric potential = 12V
Substitute these values into equation (i) as follows;
E = 3.96 x 10⁵ x 12
E = 47.52 x 10⁵ J
E = 4.752 x 10⁶ J
Therefore, the amount of energy involved is 4.752 x 10⁶ J
Ya know what the day you are talking to you about it and then send me the back link you got it and I don’t want
The force that the book exerts on the table is a normal force, not a weight force. (The book's weight doesn't act on the table, it acts on the book.) It's equal in magnitude to the weight of the book, again, because of the first law.
Answer:
0.5 m/s².
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Final velocity (v) = 10 m/s
Time (t) = 20 s
Acceleration (a) =?
Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as:
a = (v – u) / t
Where:
a is the acceleration.
v is the final velocity.
u is the initial velocity.
t is the time.
With the above formula, we can obtain the acceleration of the car as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 10 m/s
Time (t) = 20 s
Acceleration (a) =?
a = (v – u) / t
a = (10 – 0) / 20
a = 10/20
a = 0.5 m/s²
Therefore, the acceleration of the car is 0.5 m/s².
Answer:
Electric field, E = 0.064 V/m
Explanation:
It is given that,
Resistivity of silver wire, 
Current density of the wire, 
We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :


E = 0.0636 V/m
or
E = 0.064 V/m
So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.