Answer:
a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase
b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.
c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase
Explanation:
a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase
b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.
c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase
Answer:
a.) W/3, b.)2g/3 c.) (4gh/3)^0.5
Explanation:
First we have to find tension in terms of torque. To do that we have to find the moment of inertia of a rigid cylinder. From Wikipedia I get:
![I = 0.5mr^2](https://tex.z-dn.net/?f=I%20%3D%200.5mr%5E2)
We also know the equation for torque. Let T be the tension, r be the distance, I is the moment of inertia and alpha is angular acceleration (ignore theta because it is perpendicular)
![torque = I*\alpha = T*r](https://tex.z-dn.net/?f=torque%20%3D%20I%2A%5Calpha%20%3D%20T%2Ar)
We can then substitute a/r for α
Therefore we get:
![I*\frac{a}{r}=T*r](https://tex.z-dn.net/?f=I%2A%5Cfrac%7Ba%7D%7Br%7D%3DT%2Ar)
Isolating T and substitute the moment of inertia in for I we get
![T = [0.5mr^2]a/r^2= 0.5ma](https://tex.z-dn.net/?f=T%20%3D%20%5B0.5mr%5E2%5Da%2Fr%5E2%3D%200.5ma)
There are two known forces acting on the cylinder, gravity and tension. The sum of these two forces gives us mass times acceleration (Newton's second law)
![F = ma => mg - 0.5ma = ma\\g=\frac{3}{2}a\\a = \frac{2}{3}g](https://tex.z-dn.net/?f=F%20%3D%20ma%20%3D%3E%20mg%20-%200.5ma%20%3D%20ma%5C%5Cg%3D%5Cfrac%7B3%7D%7B2%7Da%5C%5Ca%20%3D%20%5Cfrac%7B2%7D%7B3%7Dg)
This allows us to plug acceleration back into Newton's Second Law:
![mg - T = m[\frac{2}{3}g]\\ T = \frac{1}{3} mg = \frac{1}{3}w](https://tex.z-dn.net/?f=mg%20-%20T%20%3D%20m%5B%5Cfrac%7B2%7D%7B3%7Dg%5D%5C%5C%20T%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20mg%20%3D%20%5Cfrac%7B1%7D%7B3%7Dw)
w = the weight
For part b, we solved in part a:
![a = \frac{2}{3}g](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B2%7D%7B3%7Dg)
For part c, we use the conservation of energy. We know that the sum of energy in the system is zero.
![mgh + \frac{1}{2} mv^2 + \frac{1}{2}I(omega)^2 = 0\\(omega) = (\frac{v}{r})^2\\ I = \frac{1}{2} mr^2\\-gh +\frac{1}{2} v^2+\frac{1}{4} v^2 = 0\\gh = \frac{3}{4} v^2\\v = \sqrt{ \frac{4}{3}gh }](https://tex.z-dn.net/?f=mgh%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DI%28omega%29%5E2%20%3D%200%5C%5C%28omega%29%20%3D%20%28%5Cfrac%7Bv%7D%7Br%7D%29%5E2%5C%5C%20I%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mr%5E2%5C%5C-gh%20%2B%5Cfrac%7B1%7D%7B2%7D%20v%5E2%2B%5Cfrac%7B1%7D%7B4%7D%20v%5E2%20%3D%200%5C%5Cgh%20%3D%20%5Cfrac%7B3%7D%7B4%7D%20v%5E2%5C%5Cv%20%3D%20%5Csqrt%7B%20%5Cfrac%7B4%7D%7B3%7Dgh%20%7D)
Answer:
microscopic light or Nano technology
Answer: The mass is 3.58*10^-2 g
Explanation: In order to calculate the mass corresponding to of a small spherical shot of copper inside a graduate cylinder with water we have considered the total volume of water incresed after put the cooper pieces.
In this sense the total increased volume of water is equal to the volume of the 125 small spherical shot of copper so we have the following:
0.5 mL= 125* volume of each piece of cooper=125*Vcopper
Vcooper=0.5/125 mL=5*10^-7 m^3/125=4*10^-9 m^3
Therefore the mass corresponding to each copper piece is equal to:
density of copper* Vcopper piece= 8960 Kg/m^3*4*10^-9 m^3=3.58*10^-5 Kg = 3.58*10^-2 g
Answer:
a)
Explanation:
- A block sliding down an inclined plane, is subject to two external forces along the slide.
- One is the component of gravity (the weight) parallel to the incline.
- If the inclined plane makes an angle θ with the horizontal, this component (projection of the downward gravity along the incline, can be written as follows:
![F_{gp} = m*g* sin \theta (1)](https://tex.z-dn.net/?f=F_%7Bgp%7D%20%3D%20m%2Ag%2A%20sin%20%5Ctheta%20%281%29)
(taking as positive the direction of the movement of the block)
- The other force, is the friction force, that adopts any value needed to meet the Newton's 2nd Law.
- When θ is so large, than the block moves downward along the incline, the friction force can be expressed as follows:
- The normal force, adopts the value needed to prevent any vertical movement through the surface of the incline:
- In equilibrium, both forces, as defined in (1), (2) and (3) must be equal in magnitude, as follows:
![m*g* sin \theta = \mu_{k} * m*g* cos \theta](https://tex.z-dn.net/?f=m%2Ag%2A%20sin%20%5Ctheta%20%3D%20%20%5Cmu_%7Bk%7D%20%2A%20m%2Ag%2A%20cos%20%5Ctheta)
- As the block is moving, if the net force is 0, according to Newton's 2nd Law, the block must be moving at constant speed.
- In this condition, the friction coefficient is the kinetic one (μk), which can be calculated as follows:
![\mu_{k} = tg \theta](https://tex.z-dn.net/?f=%5Cmu_%7Bk%7D%20%20%3D%20tg%20%5Ctheta)