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 it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.
20 J/s = 10% of Supply
20 J/s = (0.1) x (Supply)
Divide each side by 0.1:
Supply = (20 J/s) / (0.1)
<em>Supply = 200 J/s </em>(200 watts)
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Here's something to think about: What could you do to make the lamp more efficient ? Answer: Use it for a heater !
If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about. Suddenly ... bada-boom ... the lamp is 90% efficient !
Answer:
See solution below
Explanation:
The standard equation of a wave is represented by: Y = Asin 2π(x/λ – ft)
A is the amplitude
λ is the wavelength
f is the frequency
t is the time
Comparing the standard with the given equation Y = 0.005sin 2π(0.5x – 200t)
i) A = 0.005m
Amplitude is 0.005m
ii) ω = 2πf
ω = 2π(200)
ω = 400π rad/s
iii) ω = 2πf
400π = 2πf
200 = f
Swap
f = 200Hertz
iv) Period = 1/f
Period = 1/200
Period = 0.005secs
v) wave number =
vi) Wavelength λ
0.5x = x/λ
0.5 = 1/λ
λ = 1/0.5
λ = 2m
vii) velocity - fλ
velocity = 200 * 2
velocity = 400m/s
Answer:
a) K_e = 0.1225 J, b) U = 1.96 J, c) v = 0.99 m / s
Explanation:
Let's use the simple harmonium movement expression
y = A cos (wt + Ф)
indicate that the amplitude is
A = 0.05 m
as the system is released, the velocity at the initial point is zero
v = dy / dt
v = - A w sin (wt + Ф)
for t = 0 s and v = 0 m/s
0 = - A w sin Ф
so Ф = 0
the expression of the movement is
y = 0.05 cos wt
The total energy of the system is
Em = ½ k A²
let's use conservation of energy
starting point. Spring if we stretch and we set the zero of our system at this point
Em₀ = K_e + U
Em₀ = 0
final point. When weight and elastic force are in balance
Em_f = K_e + U
Em_f = ½ k y² + m g (-y)
energy is conserved
Em₀ = Em_f
0 = ½ k y² + m g (-y)
k = 2mg / y
k = 2 4.00 9.8 / 0.050
k = 98 N / m
a) maximum elastic energy
K_e = ½ k A²
K_e = ½ 98 0.05²
K_e = 0.1225 J
b) the maximum gravitational energy
U = m g y
U = 4.00 9.8 0.05
U = 1.96 J
c) The maximum kinetic energy occurs when the spring is not stretched
U = K
mg h = ½ m v²
v = √2gh
v = √( 2 9.8 0.05)
v = 0.99 m / s
d) energy at any point
Em = K + U
Answer:
If you need to measure much longer lengths - for example the length of a football pitch - then you could use a trundle wheel. You use it by pushing the wheel along the ground. It clicks every time it measures one metre.
