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4 years ago

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A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.43 m long, weighs 9.74 N, and ro

tates at 316 rev/min. Calculate (a) its rotational inertia about the axis of rotation and (b) the magnitude of its angular momentum about that axis.

1 answer:

Alecsey [184]4 years ago

8 0

The solution is in the attachment

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Vector A= 3.7 i + 1.0 j and vector B = 3.0 i + 6.5 j. What is vector<br> (A-B).A?

kicyunya [14]

Answer:

(A - B).A = -2.91

Explanation:

First, let's define the sum and dot product of vectors.

For two vectors V = (x₁, y₁) and W = (x₂, y₂) we have:

sum (or subtraction):

V + W = (x₁, y₁) + (x₂, y₂) = (x₁ + x₂, y₁ + y₂)

dot product:

V.W = (x₁, y₁).(x₂, y₂) = x₁*x₂ + y₁*y₂

Here remember the notation:

V = x₁*i + y₁*j = (x₁, y₁)

Now let's solve our problem, we have:

A = (3.7, 1.0)

B = (3.0, 6.5)

Then:

(A - B).A = ( (3.7, 1.0) - (3.0, 6.5) ).(3.7, 1.0)

= (3.7 - 3.0, 1.0 - 6.5).(3.7, 1.0)

= (0.7, -5.5).(3.7, 1.0) = (0.7*3.7) + (-5.5)*(1.0) = -2.91

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3 years ago

The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?

BigorU [14]

1. Frequency: $7.06\cdot 10^{14} Hz$

The frequency of a light wave is given by:

$f=\frac{c}{\lambda}$

where

$c=3\cdot 10^{-8} m/s$ is the speed of light

$\lambda$ is the wavelength of the wave

In this problem, we have light with wavelength

$\lambda=425 nm=425\cdot 10^{-9} m$

Substituting into the equation, we find the frequency:

$f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz$

2. Period: $1.42 \cdot 10^{-15}s$

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

The frequency of this light wave is $7.06\cdot 10^{14} Hz$ (found in the previous exercise), so the period is:

$T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s$

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