A) The answer is 11.53 m/s
The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi
Kinetic energy depends on mass (m) and velocity (v)
KEf = 1/2 m * vf²
KEi = 1/2 m * vi²
Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h
So:
KEf = KEi + <span>PEi
</span>1/2 m * vf² = 1/2 m * vi² + m * a * h
..
Divide all sides by m:
1/2 vf² = 1/2 vi² + a * h
We know:
vi = 9.87 m/s
a = 9.8 m/s²
h = 1.81 m
1/2 vf² = 1/2 * 9.87² + 9.8 * 1.81
1/2 vf² = 48.71 + 17.74
1/2 vf² = 66.45
vf² = 66.45 * 2
vf² = 132.9
vf = √132.9
vf = 11.53 m/s
b) The answer is 6.78 m
The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE
Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²
Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h
KE = PE
1/2 m * v² = m * a * h
Divide both sides by m:
1/2 * v² = a * h
v = 11.53 m/s
a = 9.8 m/s²
h = ?
1/2 * 11.53² = 9.8 * h
1/2 * 132.94 = 9.8 * h
66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m
Answer:
D
Explanation:
Because it is impossible for it to show the real depth of the ocean and how deep it is
<span>B) 0.6 N
I suspect you have a minor error in your question. Claiming a coefficient of static friction of 0.30N is nonsensical. Putting the Newton there is incorrect. The figure of 0.25 for the coefficient of kinetic friction looks OK. So with that correction in mind, let's solve the problem.
The coefficient of static friction is the multiplier to apply to the normal force in order to start the object moving. And the coefficient of kinetic friction (which is usually smaller than the coefficient of static friction) is the multiplied to the normal force in order to keep the object moving. You've been given a normal force of 2N, so you need to multiply the coefficient of static friction by that in order to get the amount of force it takes to start the shoe moving. So:
0.30 * 2N = 0.6N
And if you look at your options, you'll see that option "B" matches exactly.</span>
Answer:
![Q=3.9825\times 10^{-9} C](https://tex.z-dn.net/?f=Q%3D3.9825%5Ctimes%2010%5E%7B-9%7D%20C)
Explanation:
We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.
1 m =100 cm
Surface area =S=![\frac{360}{10000}=0.036 m^2](https://tex.z-dn.net/?f=%5Cfrac%7B360%7D%7B10000%7D%3D0.036%20m%5E2)
![\Delta d=0.8 cm=0.008 m](https://tex.z-dn.net/?f=%5CDelta%20d%3D0.8%20cm%3D0.008%20m)
![\Delta V=100 V](https://tex.z-dn.net/?f=%5CDelta%20V%3D100%20V)
We have to find the charge Q on the positive plates of the capacitor.
V=Initial voltage between plates
d=Initial distance between plates
Initial Capacitance of capacitor
![C=\frac{\epsilon_0 S}{d}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B%5Cepsilon_0%20S%7D%7Bd%7D)
Capacitance of capacitor after moving plates
![C_1=\frac{\epsilon_0 S}{(d+\Delta d)}](https://tex.z-dn.net/?f=C_1%3D%5Cfrac%7B%5Cepsilon_0%20S%7D%7B%28d%2B%5CDelta%20d%29%7D)
![V=\frac{Q}{C}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BQ%7D%7BC%7D)
Potential difference between plates after moving
![V=\frac{Q}{C_1}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BQ%7D%7BC_1%7D)
![V+\Delta V=\frac{Q}{C_1}](https://tex.z-dn.net/?f=V%2B%5CDelta%20V%3D%5Cfrac%7BQ%7D%7BC_1%7D)
![\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}](https://tex.z-dn.net/?f=%5Cfrac%7BQd%7D%7B%5Cepsilon_0S%7D%2B100%3D%5Cfrac%7BQ%28d%2B%5CDelta%20d%29%7D%7B%5Cepsilon_0S%7D)
![\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100](https://tex.z-dn.net/?f=%5Cfrac%7BQ%28d%2B%5CDelta%20d%29%7D%7B%5Cepsilon_0%20S%7D-%5Cfrac%7BQd%7D%7B%5Cepsilon_0S%7D%3D100)
![\frac{Q\Delta d}{\epsilon_0 S}=100](https://tex.z-dn.net/?f=%5Cfrac%7BQ%5CDelta%20d%7D%7B%5Cepsilon_0%20S%7D%3D100)
![\epsilon_0=8.85\times 10^{-12}](https://tex.z-dn.net/?f=%5Cepsilon_0%3D8.85%5Ctimes%2010%5E%7B-12%7D)
![Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B100%5Ctimes%208.85%5Ctimes%2010%5E%7B-12%7D%5Ctimes%200.036%7D%7B0.008%7D)
![Q=3.9825\times 10^{-9} C](https://tex.z-dn.net/?f=Q%3D3.9825%5Ctimes%2010%5E%7B-9%7D%20C)
Hence, the charge on positive plate of capacitor=![Q=3.9825\times 10^{-9} C](https://tex.z-dn.net/?f=Q%3D3.9825%5Ctimes%2010%5E%7B-9%7D%20C)