Explain what dispersion is and how astronomers use this phenomenon to study a star’s light.
1 answer:
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Answer:
1817.448 m
Explanation:
Given:
Initial acceleration = 2.24 m/s²
Time for acceleration = 10 s
From Newton's equation of motion
v = u + at
here v is the final velocity after the acceleration
v = 0 + 2.24 × 10 = 22.4 m/s
thus,
the displacement during acceleration
From Newton's equation of motion
where,
s is the distance
u is the initial speed = 0 m/s as starting from rest
a is the acceleration
t is the time
on substituting the respective values, we get
or
s₁ = 112 m
for case 2
time = 1.25 minute = 1.25 × 60 = 75 seconds
displacement, s₂ = v × t = 22.4 × 75 = 1680 m
for the case 3
acceleration = - 9.86 m/s²
now,
the final velocity for the second case is the initial velocity for this case
thus,
or
s₃ = 50.848 - 25.40 = 25.448 m
hence,
the total displacement = s₁ + s₂ + s₃ = 112 m + 1680 m + 25.448 m
= 1817.448 m
Solution:
f ( t )= 20 S ( t ) + 55/30 tS ( t )− 55/30 ( t − 30 ) S ( t − 30 )
• Taking the Laplace Transform:
F ( s ) = 20/s + 55/30 ( 1/s^2 ) – 55/30 ( 1/s^2) e^-30s = 20/s + 55/30 ( 1/s^2 ) ( 1 – e^-30s)
f ( t )= 20 S ( t ) + 55/30 tS ( t )− 55/30 ( t − 30 ) S ( t − 30 )
• Taking the Laplace Transform:
F ( s ) = 20/s + 55/30 ( 1/s^2 ) – 55/30 ( 1/s^2) e^-30s = 20/s + 55/30 ( 1/s^2 ) ( 1 – e^-30s)