(a) 6.04 rev/s
The speed of the ball is given by:
where
is the angular speed
r is the distance of the ball from the centre of the circle
In situation 1), we have
r = 0.600 m
So the speed of the ball is
In situation 2), we have
r = 0.900 m
So the speed of the ball is
So, the ball has greater speed when rotating at 6.04 rev/s.
(b) 
The centripetal acceleration of the ball is given by
where
v is the speed
r is the distance of the ball from the centre of the trajectory
For situation 1),
v = 30.6 m/s
r = 0.600 m
So the centripetal acceleration is
(c) 
For situation 2 we have
v = 34.1 m/s
r = 0.900 m
So the centripetal acceleration is
Answer:
I think is is
Explanation:
B and C why because i have a gut feeling
Answer:
L = a 1,929 10⁴ m
a = 0.1 mm = 0.1 10⁻³ m, L = 1,929 m
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
Let's use trigonometry to find the breast
tan θ = y / x
As the angles are very small
tan θ = sin θ/ cos θ = sin θ = y / x
We replace
a y / L = m λ
L = a y / m λ
The red light has a wavelength of Lam = 700 nm = 700 10⁻⁹ m, in the third pattern it is m = 3
L = a 4.05 10⁻² / (3 700 10⁻⁹)
L = a 1,929 10⁴ m
To give a specific value we must know the width of the slit, suppose a value of a = 0.1 mm = 0.1 10⁻³ m
L = 1,929 m
wave function of a particle with mass m is given by ψ(x)={ Acosαx −
π
2α
≤x≤+
π
2α
0 otherwise , where α=1.00×1010/m.
(a) Find the normalization constant.
(b) Find the probability that the particle can be found on the interval 0≤x≤0.5×10−10m.
(c) Find the particle’s average position.
(d) Find its average momentum.
(e) Find its average kinetic energy −0.5×10−10m≤x≤+0.5×10−10m.
D. The osculations show a variable rate of motion. Hope this helps:)
