Answer:
C. strike-slip fault
Explanation:
The scientist must have observed a strike- slip fault.
A fault is an evidence of brittle deformation of the crust in the presence of applied stress on earth materials. Here, the earth material is the rock subjected to tension.
Where a fault occurs, there must have been movement between two blocks of rocks. The direction of movement helps us to delineate the fault type.
- When two blocks moves past each other horizontally, it is a strike-slip fault like rubbing your palms together.
- When a block moves in the direction of the dip, it forms a dip-slip fault which results in a fault-block mountain characterized by graben and horst systems.
Option A, Plateau is a table landform usually a mountain with flat peak.
Option B is a bowl shaped stratigraphic pattern in which the youngest sequence is at the core of the strata or a fold.
So, the most fitting option is C, a strike-slip fault.
Force required to accelerate 10 kg object to 5.9 m/s/s ?
Mass = 10 kg
Acceleration = 5.9 m/s^2
Force = Mass * Acceleration
Force = 10 kg * 5.9 m/s^2
Force = 59 kg m /s^2 = 59 N
Answer:
Part(a): the capacitance is 0.013 nF.
Part(b): the radius of the inner sphere is 3.1 cm.
Part(c): the electric field just outside the surface of inner sphere is
.
Explanation:
We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '
' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as
![C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)](https://tex.z-dn.net/?f=C%20%3D%20%5Cdfrac%7B4%20%5Cpi%20%5Cepsilon_%7B0%7D%7D%7B%28%5Cdfrac%7B1%7D%7Ba%7D%20-%20%5Cdfrac%7B1%7D%7Bb%7D%29%7D~~~~~~~~~~~~~~~~~~~~~~~~~~~%281%29)
Part(a):
Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.
So the capacitance (C) of the shell is
![C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF](https://tex.z-dn.net/?f=C%20%26%3D%26%20%5Cdfrac%7BQ%7D%7BV%7D%20%3D%20%5Cdfrac%7B3%20%5Ctimes%2010%5E%7B-90%7D~C%7D%7B230~V%7D%20%3D%201.3%20%5Ctimes%2010%5E%7B-11%7D~F%20%3D%200.013~nF)
Part(b):
Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,
![&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm](https://tex.z-dn.net/?f=%26%26%20%5Cdfrac%7B1%7D%7Ba%7D%20%3D%20%5Cdfrac%7B1%7D%7Bb%7D%20%2B%20%5Cdfrac%7B1%7D%7BC%2F4%20%5Cpi%20%5Cepsilon_%7B0%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7B0.043%7D%20%2B%20%5Cdfrac%7B1%7D%7B1.3%20%5Ctimes%2010%5E%7B-11%7D%20%5Ctimes%209%20%5Ctimes%2010%5E%7B9%7D%7D%20%3D%2031.79%5C%5C%26or%2C%26%20a%20%3D%20%5Cdfrac%7B1%7D%7B31.79%7D~m%20%3D%200.031~m%20%3D%203.1~cm)
Part(c):
If we apply Gauss' law of electrostatics, then
![&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}](https://tex.z-dn.net/?f=%26%26%20E~4%20%5Cpi%20a%5E%7B2%7D%20%3D%20%5Cdfrac%7BQ%7D%7B%5Cepsilon_%7B0%7D%7D%5C%5C%26or%2C%26%20E%20%3D%20%5Cdfrac%7BQ%7D%7B4%20%5Cpi%20%5Cepsilon_%7B0%7Da%5E%7B2%7D%7D%5C%5C%26or%2C%26%20E%20%3D%20%5Cdfrac%7B3%20%5Ctimes%2010%5E%7B-9%7D%20%5Ctimes%209%20%5Ctimes%2010%5E%7B9%7D%7D%7B0.031%5E%7B2%7D%7D~N~C%5E%7B-1%7D%5C%5C%26or%2C%26%20E%20%3D%202.81%20%5Ctimes%2010%5E%7B4%7D~N~C%5E%7B-1%7D)
is iron and aluminium is there
Friction force is when you rub 2 things together and they get warm. Motion, on the other hand, is if your walking along the sidewalk - you hardly get warmer -------
Unless it's a colder day outside and you're walking SO you decide to rub your hands together to get warm, but if you were just walking , its motion and only motion - no friction :):)