Answer:
3.6μF
Explanation:
The charge on the capacitor is defined by the formula
q = CV
because the charge will be conserved
q₁ = C₁V₂
q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same
q = q₁ + q₂ = C₁V₂ + C₂V₂
CV = CV₂ + C₂V₂
CV - CV₂ = C₂V₂
C ( V - V₂) = C₂V₂
C ( V/ V₂ - V₂ /V₂) = C₂
C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF
Answer:
Explanation:
Given parameters:
Weight of object = 49N
Force applied = 12N
Unknown:
Acceleration of object = ?
Solution:
The acceleration of the object is found by dividing the force by the weight;
Acceleration =
= 0.25m/s²
Answer: 71.93 *10^3 N/C
Explanation: In order to calculate the electric field from long wire we have to use the Gaussian law, this is:
∫E*dr=Q inside/εo Q inside is given by: λ*L then,
E*2*π*r*L=λ*L/εo
E= λ/(2*π*εo*r)= 4* 10^-6/(2*3.1415*8.85*10^-12*2 )= 71.93 * 10^3 N/C
Answer:
The coefficient of performance for the cycle is 2.33.
Explanation:
Given that,
Output energy 
Work done 
We need to calculate the coefficient of performance
Using formula of the coefficient of performance

We need to calculate the 

Put the value into the formula



Now put the value of
into the formula of COP


Hence, The coefficient of performance for the cycle is 2.33.