Answer:
The bomb will remain in air for <u>17.5 s</u> before hitting the ground.
Explanation:
Given:
Initial vertical height is, 
Initial horizontal velocity is, 
Initial vertical velocity is, 
Let the time taken by the bomb to reach the ground be 't'.
So, consider the equation of motion of the bomb in the vertical direction.
The displacement of the bomb vertically is 
Acceleration in the vertical direction is due to gravity, 
Therefore, the displacement of the bomb is given as:
So, the bomb will remain in air for 17.5 s before hitting the ground.
Answer:
zero or 2π is maximum
Explanation:
Sine waves can be written
x₁ = A sin (kx -wt + φ₁)
x₂ = A sin (kx- wt + φ₂)
When the wave travels in the same direction
Xt = x₁ + x₂
Xt = A [sin (kx-wt + φ₁) + sin (kx-wt + φ₂)]
We are going to develop trigonometric functions, let's call
a = kx + wt
Xt = A [sin (a + φ₁) + sin (a + φ₂)
We develop breasts of double angles
sin (a + φ₁) = sin a cos φ₁ + sin φ₁ cos a
sin (a + φ₂) = sin a cos φ₂ + sin φ₂ cos a
Let's make the sum
sin (a + φ₁) + sin (a + φ₂) = sin a (cos φ₁ + cos φ₂) + cos a (sin φ₁ + sinφ₂)
to have a maximum of the sine function, the cosine of fi must be maximum
cos φ₁ + cos φ₂ = 1 +1 = 2
the possible values of each phase are
φ1 = 0, π, 2π
φ2 = 0, π, 2π,
so that the phase difference of being zero or 2π is maximum
Answer with Explanation:
We are given that
Mass of rock=m
Maximum height=h
a.At maximum height, velocity,v=0
We know that
Height,h=h/4
Again,
Where 
b.When height,h=3h/4
Answer:
Half life of S = 3.76secs
Explanation:
The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.
For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564
for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615
Simply say = 1/3.9 + 1/104 = 0.266015
So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active
