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
If a constant force is applied on a body, the body moves with constant acceleration.
Answer:
1.25 m
0.5 s
Explanation:
Given:
v₀ = 5 m/s
v = 0 m/s
a = -10 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (5 m/s)² + 2 (-10 m/s²) Δy
Δy = 1.25 m
Find: t
v = at + v₀
(0 m/s) = (-10 m/s²) t + (5 m/s)
t = 0.5 s
Area is a region bounded by a CLOSED path, curved path or path with line segments. Area is usually a two dimensional physical quantity, a
designated part on a plane. An example the surface of a table, face of a
wall or a ceiling. It can also be a surface in 3-dimensions. Its units are meter² or centimeter² or feet² etc. A large agricultural field is a good example of an area. Normally and ideally no object can reside or exist with in an area. An object requires a volume. It is a collection of all points enclosed or bounded by the boundary or closed path. It does not have any thickness.
Volume is a 3-dimensional quantity. A physical body occupies space with its material components. It is the amount of space occupied, that is its volume. Any existing body occupying some space has a mass and volume. A mass having a volume, has surface area and thickness, a height perpendicular to its surface at each point on the surface.
It is expressed as meter³, feet³ etc.
Volume is a product of surface area (cross-section) and length - for example for a wire or rod.
