Answer:
(a) A = 0.700m
(b) k = 80.6N/m
(c) x = -0.699m
(d) x = -0.350m
(e) t = 0.168s
Explanation:
Given the equation of motion for the spring
X = 0.700cos(12.0t), m = 0.56kg
(a) A = amplitude = 0.700m
(b) The angular velocity ω = 12rad/s
ω = √(k/m)
ω² = k/m
k = m×ω² = 0.56×12² = 80.6N/m
Spring constant k = 80.6N/m
(c) T = 2π/ω = 2π/12
T = 0.524s
At t = T/2 = 0.524/2 = 0.262s
So x = 0.700cos(12×0.262) = –0.699m
(d) At t = 2/3×T = 2×0.524/3 = 0. 349s
x = 0.700cos(12×0.349) = –0.350m
(e) to find t at x = -0.300m
–0.300 = 0.700cos(12t)
–0.300/0.700 = cos(12t)
cos(12t) = –0.429
12t = cos-¹(-0.429)
12t = 2.01
t = 2.01/12
t = 0.168s
Answer: the object should be overcome by buoyancy and rise in the fluid.
Explanation:
Answer:
0.37 m
Explanation:
Given :
Window height, 
= 1.27 m
The flowerpot falls 0.84 m off the window height, i.e.
= (1.27 x 0.84 ) m in a time span of 
seconds.
Assuming that the speed of the pot just above the window is v then,
![$v=\left(\frac{30}{8}\right) \left[ (1.27 \times 0.84) - \left( \frac{1}{2} \times 9.81 \times \left( \frac{8}{30 \right)^2 \right) \right]}$](https://tex.z-dn.net/?f=%24v%3D%5Cleft%28%5Cfrac%7B30%7D%7B8%7D%5Cright%29%20%5Cleft%5B%20%281.27%20%5Ctimes%200.84%29%20-%20%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%209.81%20%5Ctimes%20%5Cleft%28%20%5Cfrac%7B8%7D%7B30%20%5Cright%29%5E2%20%5Cright%29%20%5Cright%5D%7D%24)
m/s
Initially the pot was dropped from rest. So, u = 0.
If it has fallen from a height of h above the window then,
h = 0.37 m
Answer:
Work is the energy transferred to or from an object via the application of force along a displacement.
