Answer:
D 40,000nm
Step-by-step explanation: 1nm/10^-9m * 10^-6/1um = 10^3 nm/um
Therefore in 40 microns there are:
40um * 10^3 nm/um = 40 * 10^3 nm = 40,000 nm
Answer:
8
Step-by-step explanation:
the length of segment AD would be 8 since segment AD and segment WZ are equal in length!
Given that

we first differentiate with respect to t to get the tangent vector,
:

At t = 0, the tangent vector is

To get the <em>unit</em> tangent vector, multiply this by 1/(norm of tangent vector) :

Then the unit tangent vector is

Oh okie so I can see you in the first week or so
Answer:
V = 63π / 200 m^3
Step-by-step explanation:
Given:
- The function y = f(x) is revolved around the x-axis over the interval [1,6] to form a spherical surface:
y = √(42*x - x^2)
- The surface is coated with paint with uniform layer thickness t = 1.5 mm
Find:
The volume of paint needed
Solution:
- Let f be a non-negative function with a continuous first derivative on the interval [1,6]. The Area of surface generated when y = f(x) is revolved around x-axis over the interval [1,6] is:
![S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx](https://tex.z-dn.net/?f=S%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5Ea_b%20%7B%20%5Bf%28x%29%2A%5Csqrt%7B1%20%2B%20f%27%28x%29%5E2%7D%20%7D%5D%20%5C%2C%20dx)
- The derivative of the function f'(x) is as follows:

- The square of derivative of f(x) is:

- Now use the surface area formula:
![S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi](https://tex.z-dn.net/?f=S%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20%5B%5Csqrt%7B42x-x%5E2%7D%20%2A%5Csqrt%7B1%20%2B%20%5Cfrac%7B%2821-x%29%5E2%7D%7B42x-x%5E2%20%7D%20%7D%5D%20%5C%2C%20dx%5C%5C%5C%5CS%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20%5B%5Csqrt%7B42x-x%5E2%2B%2821-x%29%5E2%7D%20%7D%5D%20%5C%2C%20dx%5C%5C%5C%5CS%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20%5B%5Csqrt%7B42x-x%5E2%2B441-42x%2Bx%5E2%7D%20%7D%5D%20%5C%2C%20dx%5C%5C%5C%5CS%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20%5B%5Csqrt%7B441%7D%20%7D%5D%20%5C%2C%20dx%5C%5CS%20%3D%202%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%2021%7D%20%5C%2C%20dx%5C%5C%5C%5CS%20%3D%2042%2A%5Cpi%20%5Cint%5Climits%5E6_1%20%7B%20dx%7D%20%5C%2C%5C%5C%5C%5CS%20%3D%2042%2A%5Cpi%20%5B%206%20-%201%20%5D%5C%5C%5C%5CS%20%3D%2042%2A5%2A%5Cpi%20%5C%5C%5C%5CS%20%3D%20210%5Cpi)
- The Volume of the pain coating is:
V = S*t
V = 210*π*3/2000
V = 63π / 200 m^3