The work done in lifting the hamburger is equal to the increase in gravitational potential energy of the hamburger, given by

where
m=0.1 kg is the mass of the hamburger
is the gravitational acceleration
is the increase in height of the hamburger
Substituting numbers into the equation, we find

So, the correct answer is
(3) 0.3 J
Answer:
Option 3: -48 cm
Explanation:
We are given:
refractive index; n = 1.5
radius of curvature; r2 = 24 cm
Formula for the focal length is given as;
1/f = (n - 1) × [(1/r1) - (1/r2)]
As r1 tends to infinity, 1/r1 = 0
Thus,we now have;
1/f = (n - 1) × (-1/r2)
Plugging in the relevant values;
1/f = (1.5 - 1) × (-1/24)
1/f = -0.02083333333
f = -1/0.02083333333
f = -48 cm
Complete Question
A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?
Answer:
The values is 
Explanation:
From the question we are told that
The diameter is 
The charge is 
The distance from the center is 
Generally the radius is mathematically represented as

=> 
=> 
Generally electric field is mathematically represented as
![E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7BQ%7D%7B%202%5Cepsilon_o%20%7D%20%5B1%20-%20%5Cfrac%7Bk%7D%7B%5Csqrt%7Br%5E2%20%2B%20%20k%5E2%20%7D%20%7D%20%5D)
substituting values
![E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B4.4%20%2A10%5E%7B-9%7D%7D%7B%202%2A%20%288.85%2A10%5E%7B-12%7D%29%20%7D%20%5B1%20-%20%5Cfrac%7B%281.00%20%2A10%5E%7B-4%7D%29%7D%7B%5Csqrt%7B%280.06%29%5E2%20%2B%20%20%281.0%2A10%5E%7B-4%7D%29%5E2%20%7D%20%7D%20%5D)
