A spring with a mass of 5 Kg has natural length 0.5m. A force of 35.6 N is required to maintain it stretched to a length of 0.5m
1 answer:
Answer:
Explanation:
force constant of spring k = force / extension
= 35.6 / 0.5
k = 71.2 N / m
angular frequency ω of oscillation by spring mass system
where m is mass of the body attached with spring
Putting the values
ω = 3.77 radian / s
The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2
so the equation for displacement from equilibrium position that is middle point can be given as follows
x = .5 sin ( ω t + π / 2 )
= 0.5 cos ω t
= 0.5 cos 3.77 t .
x = 0.5 cos 3.77 t .
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The kinetic energy of the electrons depends on the frequency of the light:
K = h(f - f₀)
where:
K = kinetic energy
h = Planck constant
f = light frequency
f₀ = threshold frequency
Rearranging the formula in the form y = m·x + b, we get:
K = hf - hf₀
where:
K = dependent variable
f = <span>indipendent variable
h = slope
hf</span>₀ = y-intercept
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Hence, the correct answer is C) the slope.