Answer:
1) P₁ = -2 D, 2) P₂ = 6 D
Explanation:
for this exercise in geometric optics let's use the equation of the constructor
where f is the focal length, p and q are the distance to the object and the image, respectively
1) to see a distant object it must be at infinity (p = ∞)
q = f₁
2) for an object located at p = 25 cm
We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm
we substitute in the equations
1) f₁ = -50 cm
2)
= 0.06
f₂ = 16.67 cm
the expression for the power of the lenses is
P =
where the focal length is in meters
1) P₁ = 1/0.50
P₁ = -2 D
2) P₂ = 1 /0.16667
P₂ = 6 D
V = I * R
Where V is the voltage, I is the current and R is the resistance. Using Ohm's law, you require resistance to find the current through the wire. Technically, if the wire has a resistance of 0, you will get infinite current. But this isn't possible. Maybe the negligible resistance refers to the battery's internal resistance - not the wire's resistance.
Answer:
Explanation:
Force on a moving charge is given by the following relation
F = q ( v x B )
for proton
q = e , v = vi , B = Bk
F = e ( vi x Bk )
= Bev - j
= - Bevj
The direction of force is along negative of y axis or -y - axis.
for electron
q = - e , v = vi , B = Bk
F = - e ( vi x Bk )
= - Bev - j
= Bevj
The direction of force is along positive of y axis or + y - axis.
Answer:
C
Explanation:
light can travel in a vacuum Anne the sped varies