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GarryVolchara [31]
3 years ago
10

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

motion that occurs has a maximum speed of 1.7 m/s. Determine the amplitude A of the motion.
Physics
1 answer:
Umnica [9.8K]3 years ago
7 0

Answer:

<em>A = 6.9 cm</em>

Explanation:

<u>Simple Harmonic Motion</u>

A mass-spring system is a common example of a simple harmonic motion device since it keeps oscillating when the spring is stretched back and forth.

If a mass m is attached to a spring of constant k and they are set to oscillate, the angular frequency of the motion is

\displaystyle w=\sqrt{\frac{k}{m}}

The equation for the motion of the object is written as a sinusoid:

\displaystyle X=A\ cos\ w\ t

Where A is the amplitude.

The instantaneous speed is computed as the derivative of the distance

\displaystyle X'=V=-A\ w\ sin\ w\ t

And the maximum speed is

\displaystyle V_{max}= A\ w

Solving for the amplitude

\displaystyle A= \frac{V_{max}}{w}

Computing w

\displaystyle w =\sqrt{\frac{120}{0.2}}=24.5\   rad/ s

Calculating A

\displaystyle A=\frac{1.7}{24.5}=0.069\ m

\displaystyle \boxed{A=6.9\ cm}

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natali 33 [55]

Answer:

Final speed of the bullet is 228.3 m/s

Explanation:

As we know that there is no external force on the system of wooden block and the bullet

so we can say momentum of the system is conserved here

so here we can say

P_i = P_f

m_1v_1 = m_1v_{1f} + m_2v_{2f}

4.67\times 10^{-3}(613) = 4.67 \times 10^{-3}v_{1f} + (0.078)(23)

so we will have

2.86 = 4.67\times 10^{-3} v_{1f} + 1.794

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3 years ago
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3 years ago
1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp
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Answer with explanation:

We are given that  

Mass of ball,m_1=75 g=\frac{75}{1000}=0075kg

1 kg=1000 g

Height,h_1=1.6 m

h_2=0.6 m

Horizontal velocity,v_x=2 m/s

Mass of platem_2=400 g=\frac{400}{1000}=0.4 kg

a.Initial velocity of plate,u_2=0

Velocity before impact=u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s

Where g=9.8 m/s^2

Velocity after impact,v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s

According to law of conservation of momentum  

m_1u_1+m_2u_1=-m_1v_1+m_2v_2

Substitute the values  

0.075\times 5.6+0=-0.075\times 3.4+0.4v_2

0.4v_2=0.075\times 5.6+0.075\times 3.4

v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s

Velocity of plate=1.69 m/s

b.Initial energy=\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J

Final energy=\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2

Final energy=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

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A ball is shot straight up into the air from the ground with initial velocity of 44ft/sec44ft/sec. assuming that the air resista
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Convert the given in SI units.

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The distance traveled and the initial velocity can be related through the equation,
  
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The value of d from the equation,

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Convert this to feet,

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