When visible light, X rays, gamma rays, or other forms of electromagnetic radiation are shined on certain kinds of matter, electrons are ejected. That phenomenon is known as the photoelectric effect. The photoelectric effect was discovered by German physicist Heinrich Hertz (1857–1894) in 1887. You can imagine the effect as follows: Suppose that a metal plate is attached by two wires to a galvanometer. (A galvanometer is an instrument for measuring the flow of electric current.) If light of the correct color is shined on the metal plate, the galvanometer may register a current. That reading indicates that electrons have been ejected from the metal plate. Those electrons then flow through the external wires and the galvanometer. HOPE THIS HELPED
Answer:
(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).
(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .
Explanation:
m= 3kg
a= 2 i + 5 j = 5 .38 < 68.19 º
F= m * a
F= 3* ( 5.38 < 68.19º )
F= 16.4 N < 68.19º
Fx= F * cos(68.19º)
Fx= 5.99
Fy= F* sin(68.19º)
Fy= 14.98
Answer:
a) K = 0.63 J, b) h = 0.153 m
Explanation:
a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is
w² =
where d is the distance from the pivot point to the center of mass and I is the moment of inertia.
The rod is a homogeneous body so its center of mass is at the geometric center of the rod.
d = L / 2
the moment of inertia of the rod is the moment of a rod supported at one end
I = ⅓ m L²
we substitute
w =
w =
w =
w = 4.427 rad / s
an oscillatory system is described by the expression
θ = θ₀ cos (wt + Φ)
the angular velocity is
w = dθ /dt
w = - θ₀ w sin (wt + Ф)
In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1
In the exercise it is indicated that at the lowest point the angular velocity is
w = 4.0 rad / s
the kinetic energy is
K = ½ I w²
K = ½ (⅓ m L²) w²
K = 1/6 m L² w²
K = 1/6 0.42 0.75² 4.0²
K = 0.63 J
b) for this part let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ I w²
final point. Highest point
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ I w² = m g h
½ (⅓ m L²) w² = m g h
h = 1/6 L² w² / g
h = 1/6 0.75² 4.0² / 9.8
h = 0.153 m
Answer:
A concave mirror is used as a torch reflector. ... When a light bulb is placed at the focus of a concave mirror reflector, the diverging light rays of the bulb are collected by the reflector. These rays are then reflected to produce a strong, parallel-sided beam of light.
Explanation:
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