Answer:
C2, C1, C4, C5 and C6 are in parallel. Therefore, we use the formula Cp = C1 + C2 + ....
Cp = C2 + C1 + C4 + C5 + C6 = ( 7 * 10 ^-3) + (18 * 10^-6) + (0.8F) + (200 * 10^-3 F) + (750 * 10^-6) = 1.008F
Now, Cp will become one capacitor and it will be aligned with C3, therefore it will now become a circuit in series.
We use the formula: 1/Cs = 1/C1 + 1/C2 + .... + ....1/Cn
Thus,
1/Cs = 1/C3 + 1/Cp
1/Cs = 1/(14 * 10^-3 F) + 1/(1.008F)
Cs = 1.4 * 10 ^-2 or if we do not round too much it will give exactly 0.0138 F
So the answer should be a)
B) The amount of work done
Answer:
A) True, B) False, C) False and D) false
Explanation:
Let's solve the problem using the law of conservation of energy to know if the statements are true or false
Let's look for mechanical energy
Initial
Emo = Ke = ½ k Dx2
Final
Em1= ½ m v12
Emo = Em1
½ k Δx2 = ½ m v₁²
v₁² = k / m Δx²
v₁ = √ k/m Δx
Now let's calculate the speed when it falls
Vfy² = Voy² - 2gy
Vfy² = - 2gy
Vf² = v₁² + vfy²
A) True v₁ = A Δx
.B) False. As there is no rubbing the mechanical energy conserves
.C) False the velocity is proportional to the square root of the height
v2y = v2 √2
. D) false promotional compression speed
Answer:
It allows electrons to flow from one part of the circuit to another
when it is closed.
Explanation:
hope this helps <3 !!
