To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are
![\phi = 76cm](https://tex.z-dn.net/?f=%5Cphi%20%3D%2076cm)
![Error (dr) = 0.5cm](https://tex.z-dn.net/?f=Error%20%28dr%29%20%3D%200.5cm)
The radius then would be
![\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm](https://tex.z-dn.net/?f=%5Cphi%20%3D%202%5Cpi%20r%20%5C%5C76cm%20%3D%202%5Cpi%20r%5C%5Cr%20%3D%20%5Cfrac%7B38%7D%7B%5Cpi%7D%20cm)
And
![\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Cphi%7D%7Bdr%7D%20%3D%202%5Cpi%20%5C%5Cd%5Cphi%20%3D%202%5Cpi%20dr%20%5C%5C0.5%20%3D%202%5Cpi%20dr)
PART A ) For the Surface Area we have that,
![A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}](https://tex.z-dn.net/?f=A%20%3D%204%5Cpi%20r%5E2%20%5C%5CA%20%3D%204%5Cpi%20%28%5Cfrac%7B38%7D%7B%5Cpi%7D%29%5E2%5C%5CA%20%3D%20%5Cfrac%7B5776%7D%7B%5Cpi%7D)
Deriving we have that the change in the Area is equivalent to the maximum error, therefore
![\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdr%7D%20%3D%204%5Cpi%20%282r%29%20%5C%5CdA%20%3D%204r%20%282%5Cpi%20dr%29)
Maximum error:
![dA = 4(\frac{38}{\pi})(0.5)](https://tex.z-dn.net/?f=dA%20%3D%204%28%5Cfrac%7B38%7D%7B%5Cpi%7D%29%280.5%29)
![dA = \frac{76}{\pi}cm^2](https://tex.z-dn.net/?f=dA%20%3D%20%5Cfrac%7B76%7D%7B%5Cpi%7Dcm%5E2)
The relative error is that between the value of the Area and the maximum error, therefore:
![\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7BA%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B76%7D%7B%5Cpi%7D%7D%7B%5Cfrac%7B5776%7D%7B%5Cpi%7D%7D)
![\frac{dA}{A} = 0.01315 = 1.31\%](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7BA%7D%20%3D%200.01315%20%3D%201.31%5C%25)
PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so
![V = \frac{4}{3} \pi (\frac{38}{\pi})^3](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%28%5Cfrac%7B38%7D%7B%5Cpi%7D%29%5E3)
![V = \frac{219488}{3\pi^2}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B219488%7D%7B3%5Cpi%5E2%7D)
Therefore the Maximum Error would be,
![\frac{dV}{dr} = \frac{4}{3} 3\pi r^2](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdr%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D%203%5Cpi%20r%5E2)
![dV = 2r^2 (2\pi dr)](https://tex.z-dn.net/?f=dV%20%3D%202r%5E2%20%282%5Cpi%20dr%29)
![dV = 4r^2 (\pi dr)](https://tex.z-dn.net/?f=dV%20%3D%204r%5E2%20%28%5Cpi%20dr%29)
Replacing the value for the radius
![dV = 4(\frac{38}{\pi})^2(0.5)](https://tex.z-dn.net/?f=dV%20%3D%204%28%5Cfrac%7B38%7D%7B%5Cpi%7D%29%5E2%280.5%29)
![dV = \frac{2888}{\pi^2} cm^3](https://tex.z-dn.net/?f=dV%20%3D%20%5Cfrac%7B2888%7D%7B%5Cpi%5E2%7D%20cm%5E3)
And the relative Error
![\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7BV%7D%20%3D%20%5Cfrac%7B%20%5Cfrac%7B2888%7D%7B%5Cpi%5E2%7D%7D%7B%20%5Cfrac%7B219488%7D%7B3%5Cpi%5E2%7D%20%7D)
![\frac{dV}{V} = 0.03947](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7BV%7D%20%3D%200.03947)
![\frac{dV}{V} = 3.947\%](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7BV%7D%20%3D%203.947%5C%25)
When driving on roads that may be slippery, do not make any sudden changes in speed or direction. Option D is correct.
<h3 /><h3>What is a slippery surface?</h3>
The slick road sign serves as a warning. When the road is wet or ice, drivers should use extra caution and reduce their speed. When the weather is bad, avoid making any rapid changes in direction.
When driving on roads that may be slippery, do not make any sudden changes in speed or direction. It may cause accident because the vehicle can lose their balance.
Hence, option D is correct.
To learn more about the slippery surface, refer to the link;
brainly.com/question/1953680
#SPJ1
Answer/Explanation:
The weight of an object is defined as the force that is exerted due to the gravitational force.
Mathematically, it can be written as :
W = m g
Where
m is the mass of the object
g is the acceleration due to gravity
Also,
We know that the value of g varies with respect to the location. At the equator, the value of g is less as compared to the poles.
The feature of an object that affects its weight are :
Mass of the object
Location of the object
How much force Earth exerts on the object
The density is determined on the steepness of the slope. The greater the density is bases upon the steepest slope. To conclude, I'd say Line A has the steepest slope therefore has the greatest density.