Given that:
k = 500 n/m,
work (W) = 704 J
spring extension (x) = ?
we know that,
Work = (1/2) k x²
704 = (1/2) × 500 × x²
x = 1.67 m
A spring stretched for 1.67 m distance.
I think:
In motion- 40
Not moving- 20
It is strong enough to penetrate through flesh but not bone so we can see if there are fractures or breaks in our skeleton
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)
