The energy stored in a capacitor is given by:
![U= \frac{1}{2}CV^2](https://tex.z-dn.net/?f=U%3D%20%5Cfrac%7B1%7D%7B2%7DCV%5E2%20)
where
U is the energy
C is the capacitance
V is the potential difference
The capacitor in this problem has capacitance
![C=3.0 \mu F = 3.0 \cdot 10^{-6} F](https://tex.z-dn.net/?f=C%3D3.0%20%5Cmu%20F%20%3D%203.0%20%5Ccdot%2010%5E%7B-6%7D%20F)
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
Answer:
Acceleration of the ship, ![a=2.14\times 10^{-7}\ m/s^2](https://tex.z-dn.net/?f=a%3D2.14%5Ctimes%2010%5E%7B-7%7D%5C%20m%2Fs%5E2)
Explanation:
It is given that,
Mass of both ships, ![m=39000\ metric\ tons=39\times 10^6\ kg](https://tex.z-dn.net/?f=m%3D39000%5C%20metric%5C%20tons%3D39%5Ctimes%2010%5E6%5C%20kg)
Distance between two ships, d = 110 m
The gravitational force between two ships is given by :
![F=G\dfrac{m^2}{d^2}](https://tex.z-dn.net/?f=F%3DG%5Cdfrac%7Bm%5E2%7D%7Bd%5E2%7D)
![F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}](https://tex.z-dn.net/?f=F%3D6.67%5Ctimes%2010%5E%7B-11%7D%5C%20Nm%5E2%2Fkg%5E2%5Ctimes%20%5Cdfrac%7B%2839%5Ctimes%2010%5E6%5C%20kg%29%5E2%7D%7B%28110%5C%20m%29%5E2%7D)
F = 8.38 N
Let a is the acceleration. Now, using second law of motion as :
![a=\dfrac{F}{m}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7BF%7D%7Bm%7D)
![a=\dfrac{8.38\ N}{39\times 10^6\ kg}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B8.38%5C%20N%7D%7B39%5Ctimes%2010%5E6%5C%20kg%7D)
![a=2.14\times 10^{-7}\ m/s^2](https://tex.z-dn.net/?f=a%3D2.14%5Ctimes%2010%5E%7B-7%7D%5C%20m%2Fs%5E2)
So, the acceleration of either ship due to the gravitational attraction of the other is
. Hence, this is the required solution.
Given :
Walk in forward direction is 30 m .
Walk in backward direction is 25 m .
To Find :
The distance and displacement .
Solution :
We know , distance is total distance covered and displacement is distance between final and initial position .
So , distance travelled is :
D = 30 + 25 m = 55 m .
Now , we first move 30 m in forward direction and then 25 m in backward direction .
So , displacement is :
D = 30 - 25 m = 5 m .
Therefore , distance and displacement covered is 55 m and 5 m respectively .
Hence , this is the required solution .
As we know that the formula of kinetic energy will be
![KE = \frac{1}{2} mv^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
now here we know that
m = 2 kg
v = 1 m/s
so from the above equation we have
![KE = \frac{1}{2}(2)(1^2)](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%282%29%281%5E2%29)
![KE = 1 J](https://tex.z-dn.net/?f=KE%20%3D%201%20J)