Please answer this correctly
1 answer:
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Answer:
<u>Use tangent:</u>
- BL/BS = tan S
- x / 144 = tan 25°
- x = 144 tan 25°
- x = 67 ft (rounded)
Answer:
The maximum height of the second pluck is greater than that of the first pluck, hence the second pluck travels further
Also the distance of the maximum height of first pluck from the player is less than the distance of the second pluck from the player hence the second pluck travels further
Step-by-step explanation:
From the question we are told that
The maximum height of the first pluck is
The height of the second height is mathematically represented as
=>
Generally at maximum height
So
=>
Here 75 ft is the horizontal distance the second pluck traveled at maximum height
So the maximum height of the second pluck is mathematically represented as
=>
So comparing the maximum height of the first and the second pluck we see that the maximum height of the second pluck is greater than that of the first pluck, hence the second pluck travels father
Answer:
a. L{t} = 1/s² b. L{1} = 1/s
Step-by-step explanation:
Here is the complete question
The The Laplace Transform of a function ft), which is defined for all t2 0, is denoted by Lf(t)) and is defined by the improper integral Lf))s)J" e-st . f(C)dt, as long as it converges. Laplace Transform is very useful in physics and engineering for solving certain linear ordinary differential equations. (Hint: think of s as a fixed constant) 1. Find Lft) (hint: remember integration by parts) A. None of these. B. O C. D. 1 E. F. -s2 2. Find L(1) A. 1 B. None of these. C. 1 D.-s E. 0
Solution
a. L{t}
L{t} = ∫₀⁰⁰
Integrating by parts ∫udv/dt = uv - ∫vdu/dt where u = t and dv/dt = and v =
and du/dt = dt/dt = 1
So, ∫₀⁰⁰udv/dt = uv - ∫₀⁰⁰vdu/dt w
So, ∫₀⁰⁰ = [
]₀⁰⁰ - ∫₀⁰⁰
∫₀⁰⁰ = [
]₀⁰⁰ - ∫₀⁰⁰
= -1/s(∞exp(-∞s) - 0 × exp(-0s)) + [
]₀⁰⁰
= -1/s[(∞exp(-∞) - 0 × exp(0)] - 1/s²[exp(-∞s) - exp(-0s)]
= -1/s[(∞ × 0 - 0 × 1] - 1/s²[exp(-∞) - exp(-0)]
= -1/s[(0 - 0] - 1/s²[0 - 1]
= -1/s[(0] - 1/s²[- 1]
= 0 + 1/s²
= 1/s²
L{t} = 1/s²
b. L{1}
L{1} = ∫₀⁰⁰
= []₀⁰⁰
= -1/s[exp(-∞s) - exp(-0s)]
= -1/s[exp(-∞) - exp(-0)]
= -1/s[0 - 1]
= -1/s(-1)
= 1/s
L{1} = 1/s