Answer:
(a) 0.42 m
(b) 20.16 N/m
(c) - 0.42 m
(d) - 0.21 m
(e) 17.3 s
Solution:
As per the question:
Mass, m = 0.56 kg
x(t) = (0.42 m)cos[cos(6 rad/s)t]
Now,
The general eqn is:
where
A = Amplitude
= angular frequency
t = time
Now, on comparing the given eqn with the general eqn:
(a) The amplitude of oscillation:
A = 0.42 m
(b) Spring constant k is given by:
Thus
(c) Position after one half period:
(d) After one third of the period:
(e) Time taken to get at x = - 0.10 m:
t = 17.3 s
Answer:
2.4 rad/s
Explanation:
= Initial angular velocity = 3.2 rad/s
= Final angular velocity of the system
= Initial angular momentum = 1.8 kgm²
= Final angular momentum = 0.6 kgm²
As there is no external torque then the angular momentum in the system is conserved
The angular velocity of the system is 2.4 rad/s
Hello, for this applicate formula:
Vf² = Vo² + 2gh
Data:
Vf = Final Velocity = ¿?
Vo = Initial veloctiy = 0 m/s
g = Gravity = 9,81 m/s²
h = Height = 5 m
Replacing according our data:
Vf² = 0 m/s² + 2(9,81 m/s²)(5 m)
Vf² = 98,1 m²/s²
Vf = √98,1 m²/s²
Vf = 9,9 m/s
The final velocity will be <u>9,9 meters per second.</u>
The density of the air it will become foggy and and become smoky
8 miles per hour
because if it is moving at 4 miles every half hour that means you have to multiply it by two to get it in miles per hour and we all know that 4 times 2 is 8 so it would be 8 miles per hour =)