The electrostatic force between two charges is inversely
proportional to the square of the distance between them.
So if you want to multiply the force by, say, ' Q ',
you need to multiply the distance by ( 1 / √Q ) .
We want to multiply the force by 16, so we need to
multiply the distance by ( 1 / √16 ) = ( 1 / 4 ) .
The distance should be changed to 1/4 of what it is now.
Hubble time in cosmology means the estimated age of the universe and the best calculation for it is T=1/H, where H is the Hubble constant
Explanation:
an object's gravitational potential energy Eg is m×g×h where:
m=mass
g=9.8m/s²
h=height relative to the closest object below it (because it cannot potentially fall through it
so Eg = 15×9.8×5=735J
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