Answer:
h = 2.64 meters
Explanation:
It is given that,
Mass of one ball, 
Speed of the first ball, 
(upward)
Mass of the other ball, 
Speed of the other ball, 
(downward)
We know that in an inelastic collision, after the collision, both objects move with one common speed. Let it is given by V. Using the conservation of momentum to find it as :
V = 7.2 m/s
Let h is the height reached by the combined balls of putty rise above the collision point. Using the conservation of energy as :
h = 2.64 meters
So, the height reached by the combined mass is 2.64 meters. Hence, this is the required solution.
Answer:
T₁= 75.25 N : Wire tension forming angle of 52° with horizontal
T₂ = 60.49 N : Wire tension forming angle of 40° with horizontal
Explanation:
We apply Newton's first law to the holiday decoration in equilibrium
Forces acting on holiday decoration:
T₁ : Wire tension forming angle of 52° with horizontal
T₂ : Wire tension forming angle of 40° with horizontal
W= m*g= 10 kg*9.8 m/s² = 98 N : weight of the decoration
∑Fx=0
T₁x -T₂x = 0
T₁x = T₂x
T₁*cos52° = T₂*cos40°
T₁= T₂*(cos40°) / (cos52°)
T₁= 1.244T₂ Equation (1)
∑Fy=0
T₁y+T₂y -W = 0
T₁*sin52° + T₂*sin40° - 98 = 0 Equation (2)
We replace T₁ of the equation (1) in the equation (2)
1.244T₂*sin52° + T₂*sin40° - 98 = 0
0.98T₂ + 0.643T₂ = 98
1.62T₂ = 98
T₂ = 98 / 1.62
T₂ = 60.49 N
We replace T₂ = 60.49 N in the Equation (1)
T₁= 1.244*60.49 N
T₁= 75.25 N
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Answer:
the moment of inertia of the wheel is is 6.245 kg.m²
Explanation:
Given;
applied force, f = 22.04 N
radius of the wheel, r = 0.34 m
angular acceleration, a = 1.2 rad/s²
The moment of inertia of the wheel is calculated as;
Therefore, the moment of inertia of the wheel is is 6.245 kg.m²
