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blondinia [14]
2 years ago
13

Which position has the least amount of kinetic energy?

Physics
1 answer:
Anna71 [15]2 years ago
8 0

the position that has least kinetic energy is option D

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Edit question A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a t
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Answer:

176.9N

Explanation:

The following data were given

wire length,L=37cm=0.37m

linear density=18g/m

tube length,=192cm=1.92m,

speed of sound,v=343m/s

Since it is an open-closed tube, the second harmonic frequency is expressed as

f_{3}=3(\frac{v}{4l} )\\f_{3}=3(\frac{343}{4*1.92})\\f_{3}=133.98Hz

The relationship between the tension,  linear density and second harmonic frequency is expressed as

f_{3}=\frac{1}{2l_{w}}\sqrt{\frac{T}{\alpha } } \\T=(f_{3}*2l_{w})^{2}\alpha \\T=(133.984*2*0.37)^{2}*18*10^{-3}\\T=176.9N

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3 years ago
Classify Use periodic table to classify the elements potassium (K) ,Bromine (Br) and argon, (Ar) according to know likely their
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Potassium-A
Bromine-B
Argon-C
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3 years ago
The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f
tensa zangetsu [6.8K]

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

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3 years ago
A bowling ball is launched from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. T
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Let y_0 be the height of the building and thus the initial height of the ball. The ball's altitude at time t is given by

y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity.

The ball reaches the ground when y=0 after t=2.9\,\mathrm s. Solve for y_0:

0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2

\implies y_0\approx16.258\,\mathrm m

so the building is about 16 m tall (keeping track of significant digits).

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2 years ago
10. A loud motorcycle passes by the front of Bill's house. What medium
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Answer: Air

Explanation:

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3 years ago
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