A converging lens can produce both real and virtual images depending on the position of the object. Explain when converging lens
es can produce virtual images and describe the image produced.
A) A virtual image is produced if the object is on the focal point, the image is upright, enlarged and on the same side of the lens as the object.
B) A virtual image is produced if the object is on the focal point, the image is inverted, enlarged and on the opposite side of the lens from the object.
C) A virtual image is produced if the object is between the focal point and the lens, the image will be upright, enlarged and on the same side of the lens
as the object.
D) A virtual image is produced if the object is between the focal point and the
lens, the image will be upright, reduced and on the opposite side of the
lens from the object.
A) A virtual image is produced if the object is on the focal point, the image is upright, enlarged and on the same side of the lens as the object.
B) A virtual image is produced if the object is on the focal point, the image is inverted, enlarged and on the opposite side of the lens from the object.
C) A virtual image is produced if the object is between the focal point and the lens, the image will be upright, enlarged and on the same side of the lens
as the object.
D) A virtual image is produced if the object is between the focal point and the
lens, the image will be upright, reduced and on the opposite side of the
lens from the object.
2 answers:
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Answer:
L₀ = L_f , K_f < K₀
Explanation:
For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.
This means that the angular momentum before and after the collision changes.
Initial instant. Before the crash
L₀ = I₀ w₀
Final moment. Right after the crash
L_f = (I₀ + mr²) w
we treat the clay sphere as a point particle
how the angular momentum is conserved
L₀ = L_f
I₀ w₀ = (I₀ + mr²) w
w = w₀
having the angular velocities we can calculate the kinetic energy
starting point. Before the crash
K₀ = ½ I₀ w₀²
final point. After the crash
K_f = ½ (I₀ + mr²) w²
sustitute
K_f = ½ (I₀ + mr²) ( w₀)²
Kf = ½ w₀²
we look for the relationship between the kinetic energy
=
K_f < K₀
we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision
Answer:
The torque on the pulley, when the system is motionless is approximately 9.81 N·m
Explanation:
The given parameters are;
The mass of the object = 2 kg
The friction between the rope and the pulley = 0
The mass of the rope, = 0.5 kg
The mass of the pulley, = 0.01 kg
The radius of the pulley, r = 0.25 m
The torque on the pulley, τ = I·α = F × D
The torque on the pulley, when the system is motionless, τ = F × D
Where;
F = The force acting on the pulley rope = The weight of the mass ≈ 2 kg × 9.81 m/s² = 19.62 N
D = The diameter of the pulley = 2×r = 2 × 0.25 m = 0.5 m
Therefore;
τ = 19.62 N × 0.5 m = 9.81 N·m
The torque on the pulley, when the system is motionless, τ ≈ 9.81 N·m.