Answer:
I think it's C I'm so sorry if I'm wrong.
Explanation:
 
        
                    
             
        
        
        
Hi there! :)
Reference the diagram below for clarification. 
1. 
We must begin by knowing the following rules for resistors in series and parallel. 
In series:

In parallel:

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules. 

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:

2.
We can use Ohm's law to solve for the current in the circuit. 

3. 
For resistors in series, both resistors receive the SAME current. 
Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A. 
In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>. 
4. 
Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage. 

 
        
             
        
        
        
Answer:
Law 1. A body continues in its state of rest, or in uniform motion in a straight line, unless acted upon by a force.
Law 2. A body acted upon by a force moves in such a manner that the time rate of change of momentum equals the force.
Law 3. If two bodies exert forces on each other, these forces are equal in magnitude and opposite in direction.
 
        
             
        
        
        
Answer:
A) 26V
Explanation:
(a) the potential difference between the plates
Initial capacitance can be calculated using below expresion
C1= A ε0/ d1
Where d1= distance between = 2.70 mm= 2.70× 10^-3 m
ε0= permittivity of space= 8.85× 10^-12 Fm^-1
A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2 
If we substitute the values we
C1= A ε0/ d1
=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3
C1=2.589 ×10^-12 F= 2.59 pF
Initial charge can be determined using below expresion
q1= C1 × V1
V1=2.589 ×10^-12 F
V1= voltage=7.90 V
If we substitute we have
q1= 2.589 ×10^-12 × 7.90
q1= 20.45×10^-12C
20.45 pC
Final capacitance can be calculated as
C2= A ε0/ d2
d2=8.80 mm= /8.80× 10^-3
7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3
C1=0.794 ×10^-12 F= 0.794 pF
Final charge= initial charge
q2=q1 (since the battery is disconnected)
q2=q1= 20.45 pC
Final potential difference
V2= q/C2
= 20.45/0.794
= 26V
 
        
             
        
        
        
Answer:
Explanation:
BMI= weight/(height × height)          ; weight in kilogram and height in metter
      = 58kg / (1.61m  × 1.61m )
      = (58/ 2.5921) kg/
      = 22.375  kg/
      ≈ 22.4 kg/