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Citrus2011 [14]
3 years ago
11

A simple harmonic oscillator consists of a block of mass 45 g attached to a spring of spring constant 240 N/m, oscillating on a

frictionless surface. If the block is given an initial velocity of 3.5 m/s starting at the equilibrium position, what is the maximum displacement of the block from its equilibrium position
Physics
2 answers:
professor190 [17]3 years ago
8 0

Answer:

x = 4.79 cm

Explanation:

given,

mass of block = 45 g

spring constant = 240 N/m

initial velocity = 3.5 m/s

maximum displacement = ?

using conservation of energy

loss of KE = grain in spring constant

\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2

mv^2 =kx^2

x = \sqrt{\dfrac{mv^2}{k}}

x = \sqrt{\dfrac{0.045\times 3.5^2}{240}}

x = √0.00229

x = 0.0479 m

x = 4.79 cm

Brrunno [24]3 years ago
4 0

Answer:

The maximum displacement ,A= 0.047 m

Explanation:

Given that

Mass, m = 45 g= 0.045 kg

spring constant ,K= 240 N/m

Initial velocity at equilibrium ,v= 3.5 m/s

Lets take the maximum displacement is A

As we know that when the mass reached at the extreme position then the velocity of the mass will become zero.

From energy conservation

\dfrac{1}{2}KA^2=\dfrac{1}{2}mv^2

KA^2=mv^2

A=\sqrt{\dfrac{mv^2}{K}}

Now by putting the values

A=\sqrt{\dfrac{0.045\times 3.5^2}{240}}

A=0.047 m

The maximum displacement ,A= 0.047 m

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Answer:

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Explanation:

We proceed as follows;

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q = pf/p-f

From the question;

p = 4.70m

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Substituting these, we have ;

q = (4.7)(0.075)/(4.7-0.075) = 0.3525/4.625 = 0.0762 = 7.62 cm

b. The image is real and inverted since the image distance is positive

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h1 = (q/p)h0

h1 = (7.62/4.70)* 1.7

h1 = 2.76 cm

d. We want to calculate the number of pixels that fit into this image

Mathematically:

n = h1/8 micro meter

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Lorico [155]

Answer:

269 m

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Explanation:

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First, find the time it takes for the package to land.  Take the upward direction to be positive.

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Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

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t = 5.98 s

Next, find the horizontal distance traveled in that time:

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²

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Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: v

v = at + v₀

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Part 3

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Find: v

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v = -58.6 m/s

6 0
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