Answer:
n = 756.25 giga electrons
Explanation:
It is given that,
If the charge on the negative plate of the capacitor,
Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :
e is the charge on electron
or
n = 756.25 giga electrons
So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.
Answer:
Al's mass is 102.92 kg
Explanation:
As there are no external forces in the horizontal direction, the horizontal net force must be zero:
As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:
where the suffix i and f means initial and final respectively.
The initial momentum will be:
But, as they are at rest, initially
So, this means:
We know that the have an combined mass of 195 kg:
.
so:
.
Now, we can use the values:
where the minus sign appears as they are moving at opposite directions
and this is the Al's mass.
Answer:
the required minimum magnitude of the force F is 21 N
Explanation:
Given the data in the question,
m = 5 kg
width = 60 cm
height = 80 cm
Let force is F represent in the image below,
so when the block about to rotate normal shifted to edge of cube
mg(w/2) = Fh
F = mg(w/2) / h
we know that g = 9.8 m/s²
we substitute
F = (5 × 9.8 ( 60/2)) / 70
F = (5 × 9.8 × 30 ) / 70
F = 1470 / 70
F = 21 N
Therefore, the required minimum magnitude of the force F is 21 N
