The answer is B. This form of magnesium chloride is not a liquid but a solid that is white and colorless.
Well, as the waves move it moves the rope as if its trying to take shape of it. Since the rope it light it will move along the ocean and the ocean will keep pushing up on the rope. (even without the waves the water is pushing the rope up so it can take its shape)
Maybe that'll help
Answer:
V₁ = √ (gy / 3)
Explanation:
For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2
Starting point
Em₀ = U₁ + U₂
Em₀ = m₁ g y₁ + m₂ g y₂
Let's place the reference system at the point where the mass m1 is
y₁ = 0
y₂ = y
Em₀ = m₂ g y = 2 m₁ g y
End point, at height yf = y / 2
= K₁ + U₁ + K₂ + U₂
= ½ m₁ v₁² + ½ m₂ v₂² + m₁ g 
+ m₂ g
Since the masses are joined by a rope, they must have the same speed
= ½ (m₁ + m₂) v₁² + (m₁ + m₂) g
= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
How energy is conserved
Em₀ =
2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2
3/2 v₁² = 2 g y -3/2 g y
3/2 v₁² = ½ g y
V₁ = √ (gy / 3)
A) We want to find the work function of the potassium. Apply this equation:
E = 1243/λ - Φ
E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function
Given values:
E = 2.93eV, λ = 240nm
Plug in and solve for Φ:
2.93 = 1243/240 - Φ
Φ = 2.25eV
B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:
E = 1243/λ - Φ
0 = 1243/λ - Φ
0 = 1243/λ - 2.25
λ = 552nm
C) We want to find the frequency associated with the threshold wavelength. Apply this equation:
c = fλ
c = speed of light in a vacuum, f = frequency, λ = wavelength
Given values:
c = 3×10⁸m/s, λ = 5.52×10⁻⁷m
Plug in and solve for f:
3×10⁸ = f(5.52×10⁻⁷)
f = 5.43×10¹⁴Hz
Since, F = k . ∆x
Therefore, k = F / ∆x = 250 / 0.2 = 1250 N/m
(ps: convert 20 cm into 0.2 m)
