lf a heavy point mass is suspended by a weightless, inextensible and perfectly flexible string from a rigid support, then this arrangement is called simple pendulum.
In practice, however, these requirements cannot be fulfilled. So we use a practical pendulum.
A practical pendulum consists of a small metallic solid sphere suspended by a fine silk thread from a rigid support. This is the practical simple pendulum which is nearest to the ideal simple pendulum.
Note :
The metallic sphere is called the bob.
When the bob is displaced slightly to one side from its mean position and released, it oscillates about its mean position in a vertical plane.
If i was feeling harsh today, I'd say the answer to your question is impossible to obtain due to the fact that photons do not emit radiation, photons ARE the radiation emitted. Though for the sake of it, here is the method...
<u>The simple method:
</u>
E=hf
therefore f=e/h
f=(3.611x10^-15) / 6.63x10^-34)
Answer: 5.45x10^18
Answer:
the answer is a. a ball is moving towards the camera faster then slower
Answer:
0.16 m
Explanation:
A rectangular gasoline tank can hold 50.0 kg of gasoline when full, and the density of gasoline is 6.8 × 10² kg/m³. We can find the volume occupied by the gasoline (volume of the tank).
50.0 kg × (1 m³/6.8 × 10² kg) = 0.074 m³
The volume of the rectangular tank is:
volume = width × length × depth
depth = volume / width × length
depth = 0.074 m³ / 0.500 m × 0.900 m
depth = 0.16 m
Answer:
h’ = 1/9 h
Explanation:
This exercise must be solved in parts:
* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy
starting point. Higher
Em₀ = U = m g h
final point. Lower, just before the crash
Em_f = K = ½ m 
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v_b = 
* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2m 0 + m v_b
final instant. Right after the crash
p_f = (2m + m) v
the moment is preserved
p₀ = p_f
m v_b = 3m v
v = v_b / 3
v = ⅓
* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together
Starting point. Lower
Em₀ = K = ½ 3m v²
Final point. Higher
Em_f = U = (3m) g h'
Em₀ = Em_f
½ 3m v² = 3m g h’
we substitute
h’= 
h’ = 
h’ = 1/9 h
