For an ideal transformer power loss is assumed to be zero
i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage
this can be written in form of equation

here we know that


![i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1](https://tex.z-dn.net/?f=i_1%20%3D%2010%20A%7B%2Ftex%5D%3C%2Fp%3E%3Cp%3Enow%20we%20will%20use%20above%20equation%3C%2Fp%3E%3Cp%3E%5Btex%5D140%2A3.5%20%3D%2010%20%2A%20V_1)

So primary coil voltage is 49 Volts
 
        
             
        
        
        
2a for example the first one,2sec. You know that every second it moves 3metres further. So 2x3=6 but you start at 0.50m so 6+0.50=6.5
        
             
        
        
        
Answer:
The speed after being pulled is 2.4123m/s
Explanation:
The work realize by the tension and the friction is equal to the change in the kinetic energy, so:
 (1)
 (1)
Where:

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.
Additionally, the kinetic energy is equal to  , so if the initial velocity
, so if the initial velocity  is equal to zero, the initial kinetic energy
 is equal to zero, the initial kinetic energy  is equal to zero.
 is equal to zero.
Then, replacing the values on the equation and solving for  , we get:
, we get:


So, the speed after being pulled 3.2m is 2.4123 m/s
 
        
             
        
        
        
In other words a infinitesimal segment dV caries the charge 
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>