Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion

Here s = h,u = 450m/s a = -g and t = t+3
Substituting

Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting

Solving both equations

So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s

Passing of B occurs at 4108.31 height.
Beginning when the bottom of the object first touches the water,
and as it descends and more and more of it goes under, the
buoyant force on it increases during that time.
As soon as the object is completely underwater, it doesn't matter
how deep under it is, the buoyant force on it remains the same.
Answer:The central spot becomes Dark
Explanation: it become dark because as the wavelength reduces,the velocity in the detector decreases, this time by 90degrees
Answer:
a. k = (1/k₁ + 1/k₂)⁻¹ b. k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹
Explanation:
Since only one force F acts, the force on spring with spring constant k₁ is F = k₁x₁ where x₁ is its extension
the force on spring with spring constant k₂ is F = k₂x₂ where x₁ is its extension
Let F = kx be the force on the equivalent spring with spring constant k and extension x.
The total extension , x = x₁ + x₂
x = F/k = F/k₁ + F/k₂
1/k = 1/k₁ + 1/k₂
k = (1/k₁ + 1/k₂)⁻¹
B
The force on spring with spring constant k₃ is F = k₃x₃ where x₃ is its extension
Let F = kx be the force on the equivalent spring with spring constant k and extension x.
The total extension , x = x₁ + x₂ + x₃
x = F/k = F/k₁ + F/k₂ + F/k₃
1/k = 1/k₁ + 1/k₂ + 1/k₃
k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹