There is a threshold frequency for each metal, and only light of a frequency higher than this threshold causes electrons to be emitted from the metal surface.
When the metal spoon comes in contact with the hot soup, heat transfer from hot soup to the spoon because of difference in temperature of the two. The kind of heat transfer represented here is conduction. In conduction, heat is transferred from a region of higher temperature to a region of lower temperature when there is a physical contact between the two.
1) 3 miles/Hour
The speed is defined as the distance covered divided by the time taken:
![v=\frac{d}{t}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bd%7D%7Bt%7D)
where
d = 1.5 mi is the distance
t = 0.5 h is the time taken
Substituting,
![v=\frac{1.5}{0.5}=3 mi/h](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B1.5%7D%7B0.5%7D%3D3%20mi%2Fh)
2) 1.34 m/s south
Velocity, instead, is a vector, so it has both a magnitude and a direction. We have:
is the displacement in meters
is the time taken in seconds
Substituting,
![v=\frac{2414 m}{1800 s}=1.34 m/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2414%20m%7D%7B1800%20s%7D%3D1.34%20m%2Fs)
And the direction of the velocity is the same as the displacement, so it is south.
Answer:
Part a)
![\rho = 3\epsilon_0 k r^2](https://tex.z-dn.net/?f=%5Crho%20%3D%203%5Cepsilon_0%20k%20r%5E2)
Part b)
![Q = 4\pi \epsilon_0kR^5](https://tex.z-dn.net/?f=Q%20%3D%204%5Cpi%20%5Cepsilon_0kR%5E5)
Explanation:
Part a)
As we know that electric field intensity due to some given charge distribution is given as
![E = kr^3 \hat r](https://tex.z-dn.net/?f=E%20%3D%20kr%5E3%20%5Chat%20r)
now electric flux through a spherical surface of radius r is given as
![\phi = E. A](https://tex.z-dn.net/?f=%5Cphi%20%3D%20E.%20A)
![\phi = kr^3(4\pi r^2)](https://tex.z-dn.net/?f=%5Cphi%20%3D%20kr%5E3%284%5Cpi%20r%5E2%29)
now by Guass law we know that
![E.A = \frac{q}{\epsilon_0}](https://tex.z-dn.net/?f=E.A%20%3D%20%5Cfrac%7Bq%7D%7B%5Cepsilon_0%7D)
![q = 4\pi \epsilon_0kr^5](https://tex.z-dn.net/?f=q%20%3D%204%5Cpi%20%5Cepsilon_0kr%5E5)
now volume charge density is given as
![\rho = \frac{q}{\frac{4}{3}\pi r^3}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7Bq%7D%7B%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3%7D)
![\rho = 3\epsilon_0 k r^2](https://tex.z-dn.net/?f=%5Crho%20%3D%203%5Cepsilon_0%20k%20r%5E2)
Part b)
Total charge inside the radius R is given as
![Q = 4\pi \epsilon_0kR^5](https://tex.z-dn.net/?f=Q%20%3D%204%5Cpi%20%5Cepsilon_0kR%5E5)
Answer:
The average force exerted by the water on the ground is 17.53 N.
Explanation:
Given;
mass flow rate of the water, m' = 135 kg/min
height of fall of the water, h = 3.1 m
the time taken for the water to fall to the ground;
![h = ut + \frac{1}{2}gt^2\\\\h = 0 + \frac{1}{2}gt^2\\\\t = \sqrt{\frac{2\times 3.1}{9.8} } \\\\t = 0.795 \ s](https://tex.z-dn.net/?f=h%20%3D%20ut%20%2B%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%5C%5C%5C%5Ch%20%3D%200%20%2B%20%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2%5C%5C%5C%5Ct%20%3D%20%5Csqrt%7B%5Cfrac%7B2%5Ctimes%203.1%7D%7B9.8%7D%20%7D%20%5C%5C%5C%5Ct%20%3D%200.795%20%5C%20s)
mass of the water;
![m = m't\\\\m = 135 \ \frac{kg}{min} \ \times \ 0.795 \ s \ \times \ \frac{1 \ \min}{60 \ s} \ = 1.789 \ kg](https://tex.z-dn.net/?f=m%20%3D%20m%27t%5C%5C%5C%5Cm%20%3D%20135%20%5C%20%5Cfrac%7Bkg%7D%7Bmin%7D%20%5C%20%5Ctimes%20%5C%200.795%20%5C%20s%20%5C%20%5Ctimes%20%5C%20%5Cfrac%7B1%20%5C%20%5Cmin%7D%7B60%20%5C%20s%7D%20%5C%20%3D%201.789%20%5C%20kg)
the average force exerted by the water on the ground;
F = mg
F = 1.789 x 9.8
F = 17.53 N
Therefore, the average force exerted by the water on the ground is 17.53 N.