Why do the graphs differ?
1 answer:
You might be interested in
#A
Mass=4.4kg
acceleration=-1.74m/s^2
Use newtons second law
#B
initial velocity=u
Final velocity=v=0
Acceleration=a=-1.74m/s^2
Time=t=1.27s
Answer:
a) k = Mg / d
, b) v = √2gh
, c) v_{f} = , d) x² + 6d x -
dh = 0
e)the spring must compress a greater distance.
Explanation:
a) when the block of mass M is placed on the spring, we have an equilibrium condition,
∑ F = 0
- W = 0
k d = Mg
k = Mg / d
b) let's use the concepts of energy to find the velocity of the block just before the collision
starting point. Position when released
Em₀ = U = m g h
lowest point. Right at the point of shock
= K = ½ m v²2
as there is no friction, energy is conserved
Em₀ = Em_{f}
mg h = ½ m v²
v = √2gh
c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2M v + M 0
final instant. Just after the shock, before the spring compression begins
p_{f} = (2M + M) v_{f}
the moment is preserved
p₀ = p_{f}
2M v = 3M v_{f}
v_{f} = ⅔ v
v_{f} =
d) now we work with the joined system after the collision, let's use the concepts of energy
starting point. After shock, before beginning spring compression
Em₀ = K = ½ (3M)
Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²
Em₀ = 4/3 M gh
final point. With the spring fully compressed
Em_f = K_e + U = ½ k x² + (3M) g x
in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved
Em₀ = Em_f
4/3 M g h = ½ k x² + 3M g x
½ k x² + 3Mg x - 4/3 Mgh = 0
we substitute the expression for k
(
) x² + 3Mg x -
Mgh = 0
+ 3 x -
h = 0
to find the value of the spring compression, the second degree equation must be solved
x² + 6d x - dh = 0
x = [-6d ± ] / 2
x = [-6d ± 6d ]/2
x = 3d ( -1± )
e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.
Answer:
(a) the speed of helium-neon laser light in air is 3 x 10⁸ m/s
the wavelength of helium-neon laser light in air is 721.4 nm
the frequency of helium-neon laser light in air is 415.86 THz
(b) the speed of helium-neon laser light in water is 2.26 x 10⁸ m/s
the wavelength of helium-neon laser light in water is 542.4nm
the frequency of helium-neon laser light in water is 416.67THz
(c) the speed of helium-neon laser light in glass is 2 x 10⁸ m/s
the wavelength of helium-neon laser light in glass is 480.9nm
the frequency of helium-neon laser light in glass is 415.88THz
From the results above, it can be seen that speed of the light is directly proportional to its wavelength, while the frequency of the light remained fairly constant for the different media.
Explanation:
Part (a) the speed, wavelength, and frequency of helium-neon laser light in air
Given;
wavelength of helium-neon laser light in air, λ = 721.4 nm
speed of light in air, v = 3 x 10⁸ m/s
v = f λ
where;
f is the frequency of helium-neon laser light in air
f = 415.86 THz
Part (b) the speed, wavelength, and frequency of helium-neon laser light in water
refractive index of water = 1.33
Again;
f = 416.67 THz
Part (c) the speed, wavelength, and frequency of helium-neon laser light in glass
Refractive index of glass = 1.5
Also;
f = 415.88 THz