In this problem we have the electric field intensity E:
E = 6.5 × 
newtons/coulomb
We have the magnitude of the load:
q = 6.4 × 
coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × 
meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × 
)(6.5 × )(1.2 × )
PE = 5.0 x 
joules
None of the options shown is correct.
There's no way to tell. Without seeing a diagram of the circuit,
I'll need to know much more about it than you've told me.
I don't know anything about the components or power supply
that are in the circuit, and I don't know where point ' f ' is in it.
Right now, even with the copious volume of all the available
information, no answer to your question is possible.
Answer:
a). 53.75 N and 101.92 N
b). 381.44 N and 723.25 N
Explanation:
a).
ρ
, 
, 
b).
ρ, 
, 
Use energy conservation, since no energy is lost it must be constant.
E = 0.5mv² + mgh
At release the velocity v = 0 and the height is h.
E = 0 + mgh
At impact the height h = 0 and the velocity is v.
E = 0.5mv² + 0
Since the energy E is conserved:
0.5mv² = mgh
the mass m cancels and the equation becomes:
0.5v² = gh
h = 0.5v²/g
when g = 9.81 and v = 22:
h = 24,66
B is the correct answer because an object moving ten north m/s will turn into 15m/s which as you can tell is accelerating.
