Answer:

Explanation:
The change in potential energy can be expressed as:

where K is a constant with a value of
, q1 and q2 are the charges of the proton and the electron and r is the distance between them.
The charge for the proton is
and the charge for the electron is
.
Converting r=1.0nm to m:

Replacing values:


We don't know Carter, and we don't know where he is or what
he's doing, so I'm taking a big chance speculating on an answer.
I'm going to say that if Carter is pretty much just standing there,
or, let's say, lying on the ground taking a nap, then the force of
the ground acting on him is precisely exactly equal to his weight.
Answer:
The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.
Determine Fx."

Explanation:
We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.
torque=cross product of force and position . mathematically this can be express as

Where
and the position vector

using the determinant method to expand the cross product in order to determine the torque we have
![\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C2%26-3%262%5C%5C%20F_%7Bx%7D%20%267%26-5%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C)
by expanding we arrive at

since we have determine the vector value of the toque, we now compare with the torque value given in the question

if we directly compare the j coordinate we have

Answer:
option E
Explanation:
given,
diameter = 4 mm
shutter speed = 1/1000 s
diameter of aperture = ?
shutter speed = 1/250 s
exposure time to the shutter time

N is the diameter of the aperture and t is the time of exposure
now,


inserting all the values

N₂² = 4
N₂ = 2 mm
hence , the correct answer is option E