Answer:
0.191 s
Explanation:
The distance from the center of the cube to the upper corner is r = d/√2.
When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ. The new vertical distance from the center to the corner is r cos θ.
Sum of the torques:
∑τ = Iα
Fr cos θ = Iα
(k r sin θ) r cos θ = Iα
kr² sin θ cos θ = Iα
k (d²/2) sin θ cos θ = Iα
For a cube rotating about its center, I = ⅙ md².
k (d²/2) sin θ cos θ = ⅙ md² α
3k sin θ cos θ = mα
3/2 k sin(2θ) = mα
For small values of θ, sin θ ≈ θ.
3/2 k (2θ) = mα
α = (3k/m) θ
d²θ/dt² = (3k/m) θ
For this differential equation, the coefficient is the square of the angular frequency, ω².
ω² = 3k/m
ω = √(3k/m)
The period is:
T = 2π / ω
T = 2π √(m/(3k))
Given m = 2.50 kg and k = 900 N/m:
T = 2π √(2.50 kg / (3 × 900 N/m))
T = 0.191 s
The period is 0.191 seconds.
Answer:
Explanation:
Data provided in the question:
Height above the ground, H= 5.0m
Range of the ball, R= 20 m
Initial horizontal velocity = 
Initial vertical velocity= 
(Since ball was thrown horizontally only)
Acceleration acting horizontally, 
= 0 m/s² [ Since no acceleration acts horizontally) ]
Vertical Acceleration, 
= 9.8 m/s² (Since only gravity acts on it)
Let 't' be the time taken to reach ground
Therefore, using equations of motion, we have
Then using Equations of motion for horizontal motion,
Answer:
230 N
Explanation:
At the lowest position , the velocity is maximum hence at this point, maximum support force T is given by the branch.
The swinging motion of the ape on a vertical circular path , will require
a centripetal force in upward direction . This is related to weight as follows
T - mg = m v² / R
R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s
T = mg - mv² / R
T = 8.5 X 9.8 + 8.5 X 3.2² / .60 { R is length of hand of ape. }
T = 83.3 + 145.06
= 228.36
= 230 N ( approximately )
The answer is B. Nutrients.
