Your grade will probably go down to a D 68% or little higher than that
Answer:
The new force between the charges becomes double of the initial force.
Explanation:
The force acting between charge particles is given by :

k is electrostatic constant
r is distance between charges
If one of the charges are doubled, then, q₁ = 2q₁
The new force becomes,

So, the new force between the charges becomes double of the initial force.
Answer:
Explanation:
E = σ/ε = (F/A) / (ΔL/L)
E = (mg/(πd²/4) / (ΔL/L)
E = (4mg/(πd²) / (ΔL/L)
E = 4Lmg/(πd²ΔL)
E = 4(30.0)(90)(9.8)/(π(0.01²)0.25)
E = 1.35 x 10⁹ Pa or 1.35 GPa
Your answer will be (B) - intense pressure.
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