Answer:
a)
= 928 J
, b)U = -62.7 J
, c) K = 0
, d) Y = 11.0367 m, e) v = 15.23 m / s
Explanation:
To solve this exercise we will use the concepts of mechanical energy.
a) The elastic potential energy is
= ½ k x²
= ½ 2900 0.80²
= 928 J
b) place the origin at the point of the uncompressed spring, the spider's potential energy
U = m h and
U = 8 9.8 (-0.80)
U = -62.7 J
c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also
K = ½ m v²
K = 0
d) write the energy at two points, maximum compression and maximum height
Em₀ = ke = ½ m x²
= mg y
Emo = 
½ k x² = m g y
y = ½ k x² / m g
y = ½ 2900 0.8² / (8 9.8)
y = 11.8367 m
As zero was placed for the spring without stretching the height from that reference is
Y = y- 0.80
Y = 11.8367 -0.80
Y = 11.0367 m
Bonus
Energy for maximum compression and uncompressed spring
Emo = ½ k x² = 928 J
= ½ m v²
Emo =
Emo = ½ m v²
v =√ 2Emo / m
v = √ (2 928/8)
v = 15.23 m / s
ONEEEEEEEEEEEEEEEEEEEEEEEEE
Answer:
Explanation:
This is case of interference in thin films
for constructive interference in thin film the condition is
2μ t = (2n+1)λ/2 ; μ is refractive index of oil , t is thickness of oil , λ is wave length of light .
2 x 1.28 x t = λ/2 , if n = 0
2 x 1.28 x t = 605 /2
t = 118.16 nm .
the minimum non-zero thickness of the oil film required = 118.16 nm.
Build a filter.
Dig water from the ground
Hey there!
So we know that m*v=P.
And in this question m=30
v=5 m/s
P = 30*5 Kgm/s
P = 150 Kgm/s
So, your final answer is 150 Kg.m/s
Hope this helps! :)