Answer:
a) W = 180.87 J
, b) ΔU = -180.87 J
, c) ΔU = -180.87 J
Explanation:
a) Work is defined as
W = F .ds
Where bold indicates vectors, we can write the scalar product
W = F s cos θ
Where the angle is between force and displacement.
The force of gravity is the weight of the body, which is directed downwards and the displacement thickens the tip of the cliff at the bottom, so that it is directed downwards, therefore the angle is zero degrees
W = y
W = m g y
For this problem we must fix a reference system, from the statement it is established that the system is placed at the base of the cliff, so that final height is zero and the initial height (y₀ = 7.69m)
W = 2.40 9.8 (7.69-0)
W = 180.87 J
b) The potential energy is
U = mg y
The change in potential energy,
ΔU = - U₀
ΔU = mg (- y₀)
ΔU = 2.4 9.8 (0 -7.69)
ΔU = -180.87 J
c) in this case we change the reference system to the height of the cliffs, for this configuration
y₀ = 0
= -7.69 m
ΔU = 2.4 9.8 (-7.69 -0)
ΔU = -180.87 J
Answer:
at the Equator
Explanation:
The four seasons are determined by four main positions in the Earth's orbit in its turn around the Sun (ecliptic plane), which are called solstices and equinoxes: winter solstice (Capricorn point, December 22), spring equinox (Aries point, around March 21-22), summer solstice (Cancer point, June 21) and autumn equinox (Libra point, around September 22-23).
In the equinoxes, the axis of rotation of the Earth is perpendicular to the sun's rays, which fall vertically over the equator. In solstices, the axis is inclined 23.5º, so that the sun's rays fall vertically on the Tropic of Cancer (summer in the northern hemisphere) or Capricorn (summer in the southern hemisphere).
When falling vertically on Ecuador, it generates a greater impact on the surface of the Tierre reaching a greater amount of energy and therefore UV rays.
E = ½KA^2 is the mechanical energy of any oscillator. It is the sum of elastic potential energy and
kinetic energy. When amplitude A
decreases by 3%, then
(E2-E1)/E1 = {½K(A2^2/A1^2) }/ ½K(A1^2)
= {(A2^2 – A1^2) / (A1^2)}
= 97^2 – 100^2/100^2
= 5.91% of the mechanical energy is lost each cycle.