Two methods of transfer of heat are involved in this process: conduction and convection.
In fact, the metal spoon is heated by conduction because the molecules of the boiling water collide with the molecules of the spoon, releasing heat to it; and also by convection, because in the pot of boiling water masses of hot water goes upward and they give their heat to the spoon, then these masses become cooler and they go down, replaced by other masses of hot water.
Answer:
![\mu= 0.25](https://tex.z-dn.net/?f=%5Cmu%3D%200.25)
Explanation:
Given data
Force= 10N
mass= 4kg
r= 4*9.81
r= 39.24N
The expression for the force acting is expressed as
![F=\mu*r](https://tex.z-dn.net/?f=F%3D%5Cmu%2Ar)
substitute
![\mu=F/r](https://tex.z-dn.net/?f=%5Cmu%3DF%2Fr)
![\mu= 10/39.24](https://tex.z-dn.net/?f=%5Cmu%3D%2010%2F39.24)
![\mu= 0.25](https://tex.z-dn.net/?f=%5Cmu%3D%200.25)
Answer:
![\theta = 49.81^0](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2049.81%5E0)
Explanation:
Given that:
![\omega = 41.5 \ rev/min\\\\\omega = 41.5 *\frac{1}{60}* 2 \pi\\\\\omega = 4.45 \ rad/s\\\\\\diameter = 0.748 m](https://tex.z-dn.net/?f=%5Comega%20%3D%2041.5%20%5C%20rev%2Fmin%5C%5C%5C%5C%5Comega%20%3D%2041.5%20%2A%5Cfrac%7B1%7D%7B60%7D%2A%202%20%5Cpi%5C%5C%5C%5C%5Comega%20%3D%204.45%20%5C%20rad%2Fs%5C%5C%5C%5C%5C%5Cdiameter%20%3D%200.748%20m)
If we let the piece of the close lose contact at ∠θ;
Then ; from force balance;
we have:
![\\\\mg sin \theta = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})](https://tex.z-dn.net/?f=%5C%5C%5C%5Cmg%20sin%20%5Ctheta%20%3D%20%5Cfrac%7Bmv%5E2%7D%7Br%7D%5C%5C%5C%5Csin%20%5Ctheta%20%3D%20%5Cfrac%7B2v%5E2%7D%7Bdg%7D%5C%5C%5C%5C%5Ctheta%20%3D%20sin%5E%7B-1%7D%20%28%5Cfrac%7B2v%5E2%7D%7Bdg%7D%29)
where;
![v = \frac{\omega d}{2}\\\\v = \frac{4.45 *0.748}{2}\\\\v = 1.6643\\\\v^2 = 2.77](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B%5Comega%20d%7D%7B2%7D%5C%5C%5C%5Cv%20%3D%20%20%5Cfrac%7B4.45%20%2A0.748%7D%7B2%7D%5C%5C%5C%5Cv%20%3D%201.6643%5C%5C%5C%5Cv%5E2%20%3D%202.77)
Again:
![\theta = sin^{-1}(\frac{2v^2}{dg})\\\\\theta = sin^{-1}( \frac{2*2.77}{0.74*9.8})\\\\\theta = 49.81^0](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20sin%5E%7B-1%7D%28%5Cfrac%7B2v%5E2%7D%7Bdg%7D%29%5C%5C%5C%5C%5Ctheta%20%3D%20sin%5E%7B-1%7D%28%20%5Cfrac%7B2%2A2.77%7D%7B0.74%2A9.8%7D%29%5C%5C%5C%5C%5Ctheta%20%3D%2049.81%5E0)
Answer:
Yes, it is easier to climb a slanted slope than a vertical or more steep slope.
Explanation:
On a vertical slope, you are climbing higher instead of farther so on each step gravity weighs you down much more than on a gentle slope