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
The centripetal force and centripetal acceleration
both point toward the center of the circular path.
Answer:
t = 1.22 s
Explanation:
Given that,
The initial upward velocity component of a football = 12 m/s
The horizontal velocity component is 20 m/s
We need to find the time required for the football to reach the highest point of the trajectory. Let the time is t.
Using first equation of motion to solve it such that,
u is initial velocity
v is final velocity
a = -g
so,
So, the required time taken by the football to reach the highest point is 1.22 seconds.
