Answer:
0.338125 m/s
Explanation:
Applying,
Law of conservation of momentum
m'v' = mv............ Equation 1
Where m' = mass of the first skater, v' = velocity of the first skater, m = mass of the second skater, v = velocity of the second skater.
make v the subject of the equation
v = m'v'/m........... Equation 2
From the question,
Given: m' = 54.1 kg, v' = 0.375 m/s, m = 60 kg
Substitute these values into equation
v = (54.1×0.375)/60
v = 20.2875/60
v = 0.338125 m/s
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<span>node spacing = half of wavelength = 3 cm
velocity = 10 cm/s = freq * wavelength
hench freq = 10/6 = 5/3 = 1.7 hz</span>
Answer: magnitude of applied force is FA = mg + F
Where F is the resultant force downward that the rope moves with
Explanation:
Force downwards F is,
F = FA - T
T is the upwards tension force on the rope
FA is the actual applied force in pulling the rope down.
Therefore, T = FA - F .....equ. (1)
For the box to move up with force ma ( it's mass times its acceleration upwards) upwards tension on the roap must exceed its own weight mg ( it's mass times acceleration due to gravity 9.8m/s^2)
Therefore, ma = T - mg
T = ma + mg ..... equ. (2)
Equating equ. 1 and 2
T = FA - F = ma + mg
Therefore FA = ma + mg + F
But at constant velocity a = 0
Magnitude of applied force becomes
FA = mg + F
See image below
Acceleration = (final velocity - initial velocity) / time
= (35-65)/10
= -3 m/s2