As we know that fulcrum is shifted to one side by 1.33 cm
so here let say left side mass is m1 while on other side mass is m2
so we will have
now in order to find the percentage we have
so it is 10.5 % less mass
You can't tell the relative values of I-1 and I-2 from the information given,
but that's not for any super-technical reason.
The reason is because the drawing doesn't show WHERE in the circuit
I-1 and I-2 are measured.
It makes a BIG difference if they're the currents in the resistors, or
the currents in and out of the battery.
-- The currents in and out of the battery are equal.
-- The current through either resistor is less than
the current in or out of the battery.
-- The sum of the currents through both resistors
is equal to the current in or out of the battery.
-- The currents through the two resistors may be equal
or unequal, and either one may be greater or less than
the other. It all depends on the values of the resistors.
Answer:
well its simple the worker wouldn't actually be working
<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
Answer:
B. 7.07 m/s
Explanation:
The velocity of the stone when it leaves the circular path is its tangential velocity, 
, which is given by
where 
is the angular speed and 
is the radius of the circular path.
is given by
where 
is the frequency of revolution.
Thus
Using values from the question,
<em>Note the conversion of 75 cm to 0.75 m</em>
