Answer:
W=76.55 miles.metric tons
Explanation:
Given that
Weight on the earth = 12 tons
So weight on the moon =12/6 = 2 tons
( because at moon g will become g/6)
As we know that

Here x= 1100 miles
F 2 tons

So

We know that
Work = F. dx


![W=-2.4\times 10^6\left[\dfrac{1}{x}\right]_{1100}^{1140}](https://tex.z-dn.net/?f=W%3D-2.4%5Ctimes%2010%5E6%5Cleft%5B%5Cdfrac%7B1%7D%7Bx%7D%5Cright%5D_%7B1100%7D%5E%7B1140%7D)
![W=-2.4\times 10^6\left[\dfrac{1}{1140}-\dfrac{1}{1100}\right]](https://tex.z-dn.net/?f=W%3D-2.4%5Ctimes%2010%5E6%5Cleft%5B%5Cdfrac%7B1%7D%7B1140%7D-%5Cdfrac%7B1%7D%7B1100%7D%5Cright%5D)
W=76.55 miles.metric tons
In order to solve this problem, we will first need to find the electric field at the origin without the 3rd charge
E1 = (9x10^9)(13.4x10^-9)/(9.4x10^-2)^2 = 13648.7 V/m towards the negative y-axis
E2 = (9x10^9)(4.23x10^-9)/(4.99x10^-2)^2 = 15289.1 V/m towards the positive x-axis
The red arrow shows the direction of which the electric field points.
To make the electric field at the origin 0, we must find a location where q3 = the magnitude of q1 and q2
Etotal = sqrt(E1+E2) = 20494.97 V/m
E3 = 20494.97 = (9x10^9)(14.23x10^-9)/(d)^2
d = 0.079 m = 7.9 cm
Answer:
the pe at the top of the building: 784 J
the pe halfway through the fall: 392 J
the pe just before hitting the ground: 784 J
Explanation:
Pls brainliest me
I had this question before