Answer:
The equation of equilibrium at the top of the vertical circle is:
\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}
The speed experimented by the car is:
\frac{N}{m}+g=\frac{v^{2}}{R}
v = \sqrt{R\cdot (\frac{N}{m}+g) }
v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}
v\approx 9.302\,\frac{m}{s}
The equation of equilibrium at the bottom of the vertical circle is:
\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}
The normal force on the car when it is at the bottom of the track is:
N=m\cdot (\frac{v^{2}}{R}+g )
N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)
N=21.690\,N
Answer:
at the beginning: 
when the plates are pulled apart: 
Explanation:
The capacitance of a parallel-plate capacitor is given by

where
k is the relative permittivity of the medium (for air, k=1, so we can omit it)
is the permittivity of free space
A is the area of the plates of the capacitor
d is the separation between the plates
In this problem, we have:
is the area of the plates
is the separation between the plates at the beginning
Substituting into the formula, we find

Later, the plates are pulled apart to
, so the capacitance becomes

Answer:
The duration is ![T =72 \ years /tex]Explanation:From the question we are told that The distance is [tex]D = 35 \ light-years = 35 * 9.46 *10^{15} = 3.311 *10^{17} \ m](https://tex.z-dn.net/?f=T%20%20%3D72%20%5C%20%20years%20%2Ftex%5D%3C%2Fp%3E%3Cp%3EExplanation%3A%3C%2Fp%3E%3Cp%3EFrom%20the%20question%20we%20are%20told%20that%20%3C%2Fp%3E%3Cp%3E%20%20%20%20The%20%20distance%20is%20%20%5Btex%5DD%20%20%3D%20%2035%20%5C%20light-years%20%3D%2035%20%2A%20%209.46%20%2A10%5E%7B15%7D%20%3D%203.311%20%2A10%5E%7B17%7D%20%5C%20%20m%20)
Generally the time it would take for the message to get the the other civilization is mathematically represented as

Here c is the speed of light with the value 
=> 
=> 
converting to years



Now the total time taken is mathematically represented as

=> 
=> [tex]T =72 \ years /tex]
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.
The relationship between the resistance and the resistivity of a wire is

where

is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length
the cross-sectional area is given by

where r is the radius of the wire. Substituting in the previous equation ,we find

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is

Therefore, the new resistivity must be 4 times the original one.
Answer:
Well its what makes a salad taste good.