Since it is restricted at both ends, λ/2 = length of string
λ/2 = 1.5m
λ = 1.5*2 = 3m
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)
Answer:
6844.5 m/s.
Explanation:
To get the speed of the satellite, the centripetal force on it must be enough to change its direction. This therefore means that the centripetal force must be equal to the gravitational force.
Formula for centripetal force is;
F_c = mv²/r
Formula for gravitational force is:
F_g = GmM/r²
Thus;
mv²/r = GmM/r²
m is the mass of the satellite and M is mass of the earth.
Making v the subject, we have;
v = √(GM/r)
We are given;
G = 6.67 × 10^(-11) m/kg²
M = 5.97 × 10^(24) kg
r = 8500 km = 8500000
Thus;
v = √((6.67 × 10^(-11) × (5.97 × 10^(24)) /8500000) = 6844.5 m/s.