The leader Fulgencio Batista maintained control of Cuba from 1934 to 1959.
Answer: B. Fulgencio Batista
First question (upper left):
1/Req = 1/12 + 1/24 = 1/8
Req = 8 ohms
Voltage is equal through different resistors, and V1 = V2 = 24 V.
Current varies through parallel resistors: I1 = V1/R1 = 24/12 = 2 A. I2 = 24/24 = 1 A.
Second question (middle left):
V1 = V2 = 6 V (parallel circuits)
I1 = 2 A, I2 = 1 A, IT = 2+1 = 3 A.
R1 = V1/I1 = 6/2 = 3 ohms, R2 = 6/1 = 6 ohms, 1/Req = 1/2 + 1/1, Req = 2/3 ohms
Third question (bottom left):
V1 = V2 = 12 V
IT = 3 A, meaning Req = V/It = 12 V/3 A = 4 ohms
1/Req = 1/R1 + 1/R2, 1/4 = 1/12 + 1/R2, R2 = 6 ohms
I1 = V/R1 = 1 A, I2 = V/R2 = 2 A
Fourth question (top right):
1/Req = 1/20 + 1/20, Req = 10 ohms
IT = 4 A, so VT = IT(Req) = 4*10 = 40 V
Parallel circuits, so V1 = V2 = VT = 40 V
Since the resistors are identical, the current is split evenly between both: I1 = I2 = IT/2 = 2 A.
Fifth question (middle right):
1/Req = 1/5 + 1/20 + 1/4, Req = 2 ohms
IT = VT/Req = 40 V/2 ohms = 20 A
V1 = V2 = V3 = 40 V
The current of 20 A will be divided proportionally according to the resistances of 5, 20, and 4, the factors will be 5/(5+20+4), 20/(5+20+4), and 4/(5+20+4), which are 5/29, 20/29, and 4/29.
I1 = 20(5/29) = 100/29 A
I2 = 20(20/29) = 400/29 A
I3 = 20(4/29) = 80/29 A
Sixth question (bottom right):
V2 = 30V is given, but since these are parallel circuits, V1 = VT = 30 V.
Then I1 = V1/R1 = 30 V/10 ohms = 3 A.
I2 = 30 V/15 ohms = 2 A.
IT = 3 + 2 = 5 A
1/Req = 1/10 + 1/15, Req = 6 ohms
Answer:
Increasing the resistance
Explanation:
Answer
acceleration due to gravity on Jupiter's moon,g' = 1.81 m/s²
weight of water melon on earth, W = 40 N
acceleration due to gravity on earth, g = 9.8 m/s²
a) Mass on the earth surface
M = 4.08 Kg
b) Mass on the surface of Lo
Mass of an object remain same.
Hence, mass of object at the surface of Lo = 4.08 Kg.
c) Weight at the surface of Lo
W' = m g'
W' =4.08 x 1.81
W' = 7.38 N
Answer:
The total electric potential at mid way due to 'q' is 
The net Electric field at midway due to 'q' is 0.
Solution:
According to the question, the separation between two parallel plates, plate A and plate B (say) = d
The electric potential at a distance d due to 'Q' is:

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':
For plate A,
Similar is the case with plate B:
Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:


Now,
The Electric field due to charge Q at a distance is given by:

Now, if the charge q is mid way between the field, then distance is
.
Electric Field at plate A,
at midway due to charge q:

Similarly, for plate B:

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.