Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?
B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>
<span>Solution:
A) A charge q </span>under an electric field of intensity E will experience a force F equal to:
In our problem we have 
and 
, so we can find the magnitude of the electric field:
The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.
B) The proton charge is equal to
Therefore, the magnitude of the force acting on the proton will be
And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.
<span>Discrimination is illegal, but caste system is legal.
So answer: False</span>
<span>A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 115 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam.
1)What is the force the left support exerts on the beam?
2)What is the force the right support exerts on the beam?
3)How much extra mass could the gymnast hold before the beam begins to tip?
Now the gymnast (not holding any additional mass) walks directly above the right support.
4)What is the force the left support exerts on the beam?
5)What is the force the right support exerts on the beam?</span>
Answer:
Explanation:
given.
magnification(m) = 400 x
focal length (f_0)= 0.6 cm
distance between eyepiece and lens (L)= 16 cm
Near point (N) = 25 cm
focal length of the eyepiece (f_e)= ?
using equation
I think it’s D................
