Answer:
- Gap between the plates
![8.85\times 10^{- 16}\ m](https://tex.z-dn.net/?f=8.85%5Ctimes%2010%5E%7B-%2016%7D%5C%20m)
- No, practically not achievable
Solution:
As per the question:
Capacitance, C = 1 F
Area of the plate of the capacitor, A = ![1\ cm^{2} = 1\times 10^{- 4}\ m^{2}](https://tex.z-dn.net/?f=1%5C%20cm%5E%7B2%7D%20%3D%201%5Ctimes%2010%5E%7B-%204%7D%5C%20m%5E%7B2%7D)
Now,
To calculate the distance, D between the plates of a parallel plate capacitor:
![C = \frac{\epsilon_{o}A}{D}](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7B%5Cepsilon_%7Bo%7DA%7D%7BD%7D)
Thus
![D = \frac{\epsilon_{o}A}{C}](https://tex.z-dn.net/?f=D%20%3D%20%5Cfrac%7B%5Cepsilon_%7Bo%7DA%7D%7BC%7D)
where
= permittivity of free space
Now,
![D = \frac{8.85\times 10^{- 12}\times 1\times 10^{- 4}}{1}](https://tex.z-dn.net/?f=D%20%3D%20%5Cfrac%7B8.85%5Ctimes%2010%5E%7B-%2012%7D%5Ctimes%201%5Ctimes%2010%5E%7B-%204%7D%7D%7B1%7D)
![D = 8.85\times 10^{- 16}\ m](https://tex.z-dn.net/?f=D%20%3D%208.85%5Ctimes%2010%5E%7B-%2016%7D%5C%20m)
This distance much smaller and is practically not possible
Hello!
A force of 5 N produces an acceleration of 2 m/s2 on the object. What is the mass of the object ?
Data:
F (force) = 5 N
m (mass) = ?
a (acceleration) = 2 m/s²
Solving:
![F = m*a](https://tex.z-dn.net/?f=F%20%3D%20m%2Aa)
![5 = m*2](https://tex.z-dn.net/?f=5%20%3D%20m%2A2)
![2\:m = 5](https://tex.z-dn.net/?f=2%5C%3Am%20%3D%205)
![m = \dfrac{5}{2}](https://tex.z-dn.net/?f=m%20%3D%20%5Cdfrac%7B5%7D%7B2%7D)
![\boxed{\boxed{m = 2.5\:kg}}\end{array}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7Bm%20%3D%202.5%5C%3Akg%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
Answer:
2.5 kg
_______________________________
I Hope this helps, greetings ... Dexteright02! =)
Answer:
2.23 Hz
Explanation:
From the attached diagram below; there exists a diagrammatic representation of the equilibrium position of the cylinder.
The equilibrium position of the spring is expressed as:
mg = K
where;
m = mass of the object
g = acceleration due to gravity
K = spring constant
= static deflection of the string
Given that:
m = 30 kg
g = 9.81 m/s²
= 50 mm = 50 ×
= 0.05 m
Then;
![30 * 9.81= k * 0.05\\k = \frac{30*9.81}{0.05} \\k = 5886 N/m](https://tex.z-dn.net/?f=30%20%2A%209.81%3D%20k%20%2A%200.05%5C%5Ck%20%3D%20%5Cfrac%7B30%2A9.81%7D%7B0.05%7D%20%5C%5Ck%20%3D%205886%20N%2Fm)
From here; let us find the angular velocity which will be needed to determine the natural frequency aftewards.
The angular velocity of the cylinder can be expressed by the formula:
![\omega_{n} = \sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=%5Comega_%7Bn%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
![\omega_{n} = \sqrt{\frac{5886}{30}}](https://tex.z-dn.net/?f=%5Comega_%7Bn%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B5886%7D%7B30%7D%7D)
![\omega_{n} = \sqrt{196.2}](https://tex.z-dn.net/?f=%5Comega_%7Bn%7D%20%3D%20%5Csqrt%7B196.2%7D)
![\omega_{n} = 14.007141 \ \ rad/s](https://tex.z-dn.net/?f=%5Comega_%7Bn%7D%20%3D%2014.007141%20%5C%20%5C%20rad%2Fs)
Finally; the natural frequency
can be calculated by using the equation
![f_n = \frac{\omega_n}{2 \ \pi }](https://tex.z-dn.net/?f=f_n%20%3D%20%5Cfrac%7B%5Comega_n%7D%7B2%20%5C%20%5Cpi%20%7D)
![f_n = \frac{14.007141}{2 \ \pi }](https://tex.z-dn.net/?f=f_n%20%3D%20%5Cfrac%7B14.007141%7D%7B2%20%5C%20%5Cpi%20%7D)
= 2.229305729
≅ 2.23 Hz
Thus; the resulting natural frequency of the vertical vibration of the cylinder = 2.23 Hz
Answer:
Ocean water is more dense because of the salt in it. ... Temperature has a greater effect on the density of water than salinity does. So a layer of water with higher salinity can actual float on top of water with lower salinity if the layer with higher salinity is quite a bit warmer than the lower salinity layer.