Answer:
Because the disturbances are in opposite directions for this superposition, the resulting amplitude is zero for pure destructive interference
Explanation:
Answer:
Explanation:
Let the time required for acceleration a₁ and deceleration a₂ be t₁ and t₂ .
Since final velocity during acceleration and initial velocity during deceleration are same
a₁ t₁ = a₂ t₂
5t₁ = 2 t₂ ------------------------------------------ ( 1 )
Distance travelled during acceleration = 1/2 a₁t₁²
= 1/2 x 5 x t₁² = 2.5 t₁²
Distance travelled during deceleration = 1/2 a₂t₂²
= 1/2 x 2 x t₂² = t₂²
Total distance travelled = 2 miles = 2 x 1760 x 3 ft = 10560
2.5 t₁² + t₂² = 10560
2.5 ( 2t₂ / 5 )² + t₂² = 10560
.4 t₂² + t₂² = 10560
1.4 t₂² = 10560
t₂ = 86.85 s
t₁ = 2t₂ / 5 = 34.75 s
t₁ + t₂ = 121.6 = 122 s
Total time taken = 122 s .
maximum velocity = a₁t₁
= 5 x 34.75 = 173.75 = 174 m/s .
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Answer:
1.) Time t = 3.1 seconds
2.) Height h = 46 metres
Explanation:
given that the initial velocity U = 30 m/s
At the top of the trajectory, the final velocity V = 0
Using first equation of motion
V = U - gt
g is negative 9.81m/^2 as the object is going against the gravity.
Substitute all the parameters into the formula
0 = 30 - 9.81t
9.81t = 30
Make t the subject of formula
t = 30/9.81
t = 3.058 seconds
t = 3.1 seconds approximately
Therefore, it will take 3.1 seconds to reach to reach the top of its trajectory.
2.) The height it will go can be calculated by using second equation of motion
h = ut - 1/2gt^2
Substitutes U, g and t into the formula
h = 30(3.1) - 1/2 × 9.8 × 3.1^2
h = 93 - 47.089
h = 45.911 m
It will go 46 metres approximately high.
Answer:
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Explanation:
