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Vesnalui [34]
3 years ago
8

In an effort to protect a rhino, volunteers are following its steps with air monitoring and ground cameras. The rhino starts on

day 1 from ground camera A and walks for 1.5 km west along a straight line as seen from the air. The rhino then moves 0.7 km on a straight line in a direction of 15o east of north toward ground camera B. In this location the rhino may find water and food, therefore the animal stays for the night. On the second day, the rhino moves 2.5 km directly south as recorded from the air by the volunteers. In these two days of tracking how far and in which direction does the rhino travel from camera A?
Physics
1 answer:
finlep [7]3 years ago
6 0

Answer:

Explanation:

We shall represent each displacement by vectors . i will represent east , -i west , j north and - j south .

Rhino walks 1.5 km west on day 1.

D₁ = - 1.5 i

The rhino then moves 0.7 km on a straight line in a direction of 15o east of north toward ground camera B

D₂ = .7 sin15 i + .7cos15 j

On the second day, the rhino moves 2.5 km directly south

D₃ = - 2.5 j

D = D₁ + D₂ + D₃

= - 1.5 i + .7 sin15 i + .7cos15 j - 2.5 j

= - 1.5 i + .181 i + .676 j - 2.5 j

= - 1.32 i - 1.824 j

magnitude  of total displacement

= √ (1.32² +1.824²

= 2.25 km

For direction we shall calculate slope with x axis

Tanθ = - 1.824 / - 1.32

= 54°

So rhino will be towards 54° south of west as both x and y coordinates are negative.

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Answer:

r₂ = 0.316 m

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           β₁ = 10 log \frac{I_1}{I_o}

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          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

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having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

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the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

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we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

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          r₂ = \frac{10.0}{\sqrt{10^3} }

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sdas [7]
Data:
f_{2} = 42 Hz
n (Wave node)
V (Wave belly) 
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
f_{n} =  \frac{nV}{2L}

Wire 2 → 2º Harmonic → n = 2

f_{n} = \frac{nV}{2L}
f_{2} = \frac{2V}{2L} &#10;
2V =  f_{2} *2L
V =  \frac{ f_{2}*2L }{2}
V =  \frac{42*2L}{2}
V =  \frac{84L}{2}
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Wire 1 → 1º Harmonic or Fundamental rope → n = 1


f_{n} = \frac{nV}{2L}
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f_{1} =  \frac{V}{2L}

If, We have:
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f_{1} = \frac{42L}{2L}
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Answer:

<span>The fundamental frequency of the string:
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7 0
3 years ago
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3 0
3 years ago
Help me please I can't get the final step​
inna [77]

Answer:

\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

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and the dimensions of each variable is:

A=L^2T^2

B=LT^{-1}

C=LT^2

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n

Operating:

L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)

L^2T^2=L^{m+m}T^{-m+2n}

Equating the exponents:

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3n=4

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m=2-4/3=2/3

Answer:

\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}

6 0
3 years ago
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