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enyata [817]
2 years ago
9

Suppose the fetus's ventricular wall moves back and forth in a pattern approximating simple harmonic motion with an amplitude of

1.7 mm and a frequency of 3.0 Hz. Find the maximum speed of the heart wall (in m/s) during this motion. Be careful of units!
Physics
1 answer:
Luda [366]2 years ago
6 0

Answer:

The maximum speed of the heart wall during this motion is 0.032 m/s.

Explanation:

Given that,

Amplitude of the simple harmonic motion, A = 1.7 mm = 0.0017 m

Frequency of the fetus's ventricular wall, f = 3 Hz

We need to find the maximum speed of the heart wall during this motion. The maximum speed of the object that is executing SHM is given by :

v_{max}=A\omega

v_{max}=2\pi f A

v_{max}=2\pi \times 3\times 0.0017

v_{max}=0.032\ m/s

So, the maximum speed of the heart wall during this motion is 0.032 m/s. Hence, this is the required solution.

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A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu
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please read the answer below

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By taking into account the angles between the vectors r and v in each case we obtain:

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angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

b)

r=(0,-1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

c)

r=(1,0)

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r=(-1,0)

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e)

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f)

r=(-1,1)

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L = 5kgm/s

g)

r=(-1,-1)

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L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}

h)

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

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2 years ago
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