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aivan3 [116]
2 years ago
13

A 2.15 kg lightly damped harmonic oscillator has an angular oscillation frequency of 0.261 rad/s. If the maximum displacement of

1.2 m occurs when t = 0.00 s, and the damping constant b is 0.74 kg/s what is the object's displacement when t = 4.01 s?
Physics
1 answer:
Stella [2.4K]2 years ago
4 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below are the choices:

A) 0.50 m
B) 0.43 m
C) 0.58 m
D) 0.65 m
Answer: A
<span>Var: 21</span>
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Adam is teeing off on hole number two. The hole is 390 yards away. It is a par four hole. What club should he use to tee off? Ex
kkurt [141]

Answer:

A driver.

Explanation:

Using a driver while at least 350 yds away is better than using a iron, because it will be a waste of the par 4 as it is not as powerful as the driver.

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A device used to increase or decrease the emf in the second of two unconnected coils is a
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List some reasons why growth characteristics are more useful on agar plates than on agar slants
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An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of
Vsevolod [243]

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

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6 0
2 years ago
Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree tha
emmasim [6.3K]

Answer:

The distance between knothole and the paint ball is 0.483 m.

Explanation:

Given that,

Height = 4.0 m

Distance = 15 m

Speed = 50 m/s

The angle at which the forester aims his gun are,

\tan\theta=\dfrac{4}{15}

\tan\theta=0.266

\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}

\cos\theta=0.966

Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

x=(u\cos\theta)t

t=\dfrac{x}{u\cos\theta}

Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

y=u\sin\theta(t)-\dfrac{1}{2}gt^2

y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2

y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}

Put the value into the formula

y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})

y=3.517\ m

We need to calculate the distance between knothole and the paint ball

d=h-y

d=4-3.517

d=0.483\ m

Hence, The distance between knothole and the paint ball is 0.483 m.

8 0
2 years ago
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