a 30 kg child is sitting 2 meters from the center of a merry go round. The coefficients of static and kinetic friction between t
he child and the surface of the merry go round are 0.8 and 0.6 respectively. Determine the maximum speed of the merry go round before the child begins to slip. A) 
m/s B) 4 m/s c) 8 m/s D) 12 m/s E)16 m/s
1 answer:
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Answer:
Explanation:
The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2 i.e. and
The wavelength of Hi line of the Balmer series is given by :
So, the wavelength for this line is 550 nm. Hence, this is the required solution.
Answer:
A) P.E = 138.44 J
B) The velocity of swing at bottom, v = 3.33 m/s
C) The work done, W = -138.44 J
Explanation:
Given,
The mass of the child, m = 25 Kg
The length of the swing rope, L = 2.2 m
The angle of the swing to the vertical position, ∅ = 42°
A) The potential energy at the initial position ∅ = 42° is given by the relation
P.E = mgh joule
Considering h = 0 for the vertical position
The h at ∅ = 42° is h = L (1 - cos∅)
P.E = mgL (1 - cos∅)
Substituting the given values in the above equation
P.E = 25 x 9.8 x 2.2 (1 - cos42°)
= 138.44 J
The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J
B) The velocity of the swing at the bottom.
At bottom of the swing the P.E is completely transformed into the K.E
∴ K.E = P.E
1/2 mv² = 138.44
1/2 x 25 x v² 138.44
v² = 11.0752
v = 3.33 m/s
The velocity of the swing at the bottom is, v = 3.33 m/s
C) The work done by the tension in the rope from initial position to the bottom
Tension on string, T = Force acting on the swing, F
=
= - 2.2 x 25 x 9.8 [cos0 - cos 42°]
= - 138.44 J
The negative sign in the in energy is that the work done is towards the gravitational force of attraction.
The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J