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Klio2033 [76]
1 year ago
15

A displacement vector is 34.0 m in length and is directed 60.0° east of north. What are the components of this vector? Northward

Eastward choice component component 1 29.4 m 17.0 m 2 18.2 m 28.1 m 3 22.4 m 11.5 m 4 17.0 m 29.4 m 5 25.2 m 18.2 m A) Choice 1 B) Choice 2 C) Choice 3 D) Choice 4 E) Choice 5 79) If I walk 8.0 meters in a straight line and then walk 5.0 meters in another straight line, the total displacement cannot have a magnitude of A) 8 m. B) 2 m. C) 13 m. D) 5 m. E) 3 m.
Physics
1 answer:
PolarNik [594]1 year ago
8 0

Answer:

A)  29.4 m 17.0 m; B) 2 m

Explanation:

If a vector is 34.0 m in length and is directed 60.0° east of north (which means 30.0° over the horizontal), then its coordinates will be:

Horizontal: (34.0 m)cos(30.0°)=29.4 m

Vertical: (34.0 m)sin(30.0°)=17 m

If one person walks 8.0 meters in a straight line and then walks 5.0 meters in another straight line, then the minimum displacement would be to go back over his tracks, displacing himself 8m-5m=3m, while the maximum displacement would be going straight ahead, displacing himself 8m+5m=13m. Any answer outside this interval is impossible (2m).

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A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is
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Where

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Our values are given as,

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7 0
2 years ago
A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glo
padilas [110]

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Where m corresponds to the diffraction order

Let's use trigonometry to find the breast

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The diffraction spectrum is measured at very small angles, therefore

      tan θ = sin θ / cos θ = sin θ

We replace

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Let's place in the first order m = 1

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Now we can look for the wavelength of the other line

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4 0
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Answer:

Smaller refractive power

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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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