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3 years ago

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A typical helicopter has four blades that rotate at 334 rpm and have a total kinetic energy of 4.55 105 J. What is the magnitude

of the total angular momentum of the blades about the axis of rotation?

1 answer:

djyliett [7]3 years ago

5 0

Answer:

the magnitude of the total angular momentum of the blades is <em>743.71 kg·m²</em>

Explanation:

Converting the angular speed into radians per second:

ω = 334 rpm · (2π rad / 1 rev) · (1 min / 60 s)

ω = 34.98 rad/s

The rotational kinetic energy of the blades is given by:

EK = 1/2 I ω²

where

I is the moment of inertia
ω is the angular speed

Therefore, rearranging the above equation, we get:

1/2 I ω² = EK

I ω² = 2 EK

I = 2(EK) / ω²

I = 2(4.55 × 10⁵ J) / (34.98 rad/s)²

<em>I = 743.71 kg·m²</em>

<em></em>

Therefore, the magnitude of the total angular momentum of the blades is <em>743.71 kg·m²</em>.

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Suppose that the sound level of a conversation is initially at an angry 71 dB and then drops to a soothing 54 dB. Assuming that

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Answer:

Explanation:

In the decibel scale , intensity of sound changes logarithmically as follows

$10log\frac{I}{I_0} =$ Value in decibel scale , the value of I₀ = 10⁻¹² W /m².

Putting the values

$10log\frac{I}{10^{-12}} = 71$

$log\frac{I}{10^{-12}} = 7.1$

$\frac{I}{10^{-12}} = 10^{7.1}$

$I= 10^{-4.9}$ W/m²

Similarly for 54 dB sound intensity can be given as follows

I = 10⁻¹² x $10^{5.4}$

$I= 10^{-6.6 }$ W / m²

For intensity of sound the relation is as follows

I = 2π²υ²A²ρc where υ is frequency , A is amplitude , ρ is density of air and c is velocity of sound .

Putting the given values for 71 dB

$I= 10^{-4.9}$ = 2π² x 504²xA²x 1.21 x 346

A² = 60.03 x 10⁻¹⁶

A = 7.74 x 10⁻⁸ m

For 54 dB sound

$10^{-6.6}$ = 2π² x 504²xA²x 1.21 x 346

A² = 1.1978 x 10⁻¹⁶

A = 1.1 x 10⁻⁸ m

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Which of the following is a TRUE statement about the differences between electromagnetic and mechanical waves?

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' D ' is the only correct statement on the list.

Electromagnetic waves are capable of traveling through

many substances, and they're also capable of traveling

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Answer:

Explanation:

(A)

The string has set of normal modes and the string is oscillating in one of its modes.

The resonant frequencies of a physical object depend on its material, structure and boundary conditions.

The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.

Given below are the incorrect options about the wave in the string.

• The wave is travelling in the +x direction

• The wave is travelling in the -x direction

• The wave will satisfy the given boundary conditions for any arbitrary wavelength $\lambda_i$

• The wave does not satisfy the boundary conditions $y_i(0;t)=0
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Here, the string of length L held fixed at both ends, located at x=0 and x=L

The key constraint with normal modes is that there are two spatial boundary conditions, $y(0,1)=0
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and $y(L,t)=0$

.The spring is fixed at its two ends.

The correct options about the wave in the string is

• The wavelength $\lambda_i$ can have only certain specific values if the boundary conditions are to be satisfied.

(B)

The key factors producing the normal mode is that there are two spatial boundary conditions, $y_i(0;t)=0$ and $y_i(L;t)=0$ , that are satisfied only for particular value of $\lambda_i$ .

Given below are the incorrect options about the wave in the string.

• $A_i$ must be chosen so that the wave fits exactly o the string.

• Any one of $A_i$ or $\lambda_i$ or $f_i$ can be chosen to make the solution a normal mode.

Hence, the correct option is that the system can resonate at only certain resonance frequencies $f_i$ and the wavelength $\lambda_i$ must be such that $y_i(0;t) = y_i(L;t)=0
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(C)

Expression for the wavelength of the various normal modes for a string is,

$\lambda_n=\frac{2L}{n}$ (1)

When $n=1$ , this is the longest wavelength mode.

Substitute 1 for n in equation (1).

$\lambda_n=\frac{2L}{1}\\\\2L$

When $n=2$ , this is the second longest wavelength mode.

Substitute 2 for n in equation (1).

$\lambda_n=\frac{2L}{2}\\\\L$

When $n=3$ , this is the third longest wavelength mode.

Substitute 3 for n in equation (1).

$\lambda_n=\frac{2L}{3}$

Therefore, the three longest wavelengths are $2L$ , $L$ and $\frac{2L}{3}$ .

(D)

Expression for the frequency of the various normal modes for a string is,

$f_n=\frac{v}{\lambda_n}$

For the case of frequency of the $i^{th}$ normal mode the above equation becomes.

$f_i=\frac{v}{\lambda_i}$

Here, $f_i$ is the frequency of the $i^{th}$ normal mode, v is wave speed, and $\lambda_i$ is the wavelength of $i^{th}$ normal mode.

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