Complete question:
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.
Answer:
The value of its capacitance is 1.027 x 10⁻¹² F
Explanation:
Given;
area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²
separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m
voltage of the battery, V = 18 V
The value of its capacitance is calculated as;
![C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7Bk%5Cepsilon_0A%7D%7Bd%7D%20%5C%5C%5C%5CC%20%3D%20%5Cfrac%7B%281%29%288.85%5Ctimes%2010%5E%7B-12%7D%29%282.9%20%5Ctimes%2010%5E%7B-4%7D%29%7D%7B2.5%20%5Ctimes%2010%5E%7B-3%7D%7D%20%5C%5C%5C%5CC%20%3D%201.027%20%5Ctimes%2010%5E%7B-12%7D%20%5C%20F)
Therefore, the value of its capacitance is 1.027 x 10⁻¹² F
Answer:
0.94 m/s^2 downwards
Explanation:
m = 70 kg, m g = 70 x 9.8 = 686 N
R = 620 N
Let the acceleration be a, as the apparent weight decreases so the elevator is moving downwards with an acceleration a.
mg - R = ma
686 - 620 = 70 x a
a = 0.94 m/s^2
Answer:
1.08 m/s
Explanation:
This can be solved with two steps, first we need to find the time taken to fall 9.5 m, then we can divide the horizontal distance covered with time taken to calculate the velocity.
Time taken to fall 9.5 m
vertical acceleration = a = 9.8 m/s^2.
vertical velocity = 0, (since there is only horizontal component for velocity,
)
distance traveled s = 9.5 m.
Substituting these values in the equation
![s= u \timest+0.5at^{2}](https://tex.z-dn.net/?f=s%3D%20u%20%5Ctimest%2B0.5at%5E%7B2%7D)
![t= \sqrt{\frac{2s}{g} }](https://tex.z-dn.net/?f=t%3D%20%5Csqrt%7B%5Cfrac%7B2s%7D%7Bg%7D%20%7D)
![t=\sqrt{\frac{2\times9.5}{9.8} }](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cfrac%7B2%5Ctimes9.5%7D%7B9.8%7D%20%7D)
⇒ t= 1.392 sec
Velocity needed
We know the time taken (1.392 s) to travel 1.5 m,
So velocity = 1.5 m / 1.392 s = 1.08 m/s
hence velocity of the diver must be at least 1.08 m/s