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lesya692 [45]
3 years ago
15

Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y

ou from your starting point, and what is the compass direction of a line connecting your starting point to your final position?

Physics
1 answer:
Gnesinka [82]3 years ago
6 0
The representation of this problem is shown in Figure 1. So our goal is to find the vector \overrightarrow{R}. From the figure we know that:

\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}

From geometry, we know that:

\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}

Then using vector decomposition into components:

For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85

Therefore:

R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector \overrightarrow{R}, that is:
</span>
\left | \overrightarrow{R} \right |=&#10;\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span>\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}


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Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of
natita [175]

Answer:

The phase difference is 0.659 rad.

Explanation:

Given that,

Distance between two identical loudspeakers d= 1.00  m

Distance between speakers and listener r= 4.00 m

Frequency = 300 Hz

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We need to calculate the r'

Using Pythagorean theorem

r'=\sqrt{d^2+r^2}

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r = distance between speakers and listener

Put the value into the formula

r'=\sqrt{(1.00)^2+(4.00)^2}

r'=\sqrt{1.00+16.00}

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We need to calculate the path difference

Using formula of path difference

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|r'-r|=0.12\ m

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{v}{f}

Where, v = speed of sound

f = frequency

Put the value into the formula

\lambda=\dfrac{343}{300}

\lambda=1.143\ m

We need to calculate the phase difference

Using formula of phase difference

\phi=\dfrac{2\pi\times|r'-r| }{\lambda}

\phi=\dfrac{2\pi\times0.12}{1.143}

\phi=0.659\ rad

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If a particle's position is given by x=4-12t+3t^2, where t is in seconds and x is in meters, what is its velocity at t=1 second?
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v = -6m/s

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3 years ago
A parallel-plate capacitor has square plates that are 8.00cm on each side and 4.20mm apart. The space between the plates is comp
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Answer:

U=1.29\times 10^{-7}\ J

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d₂=2.1 mm  , K₂=2.6

We know that capacitance given as

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U=\dfrac{CV^2}{2}

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3 years ago
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