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konstantin123 [22]

2 years ago

8

of 72.0 units. If the mass of Object 1 is halved AND the mass of object 2 is tripled, then the new gravitational force will be u

nits.

1 answer:

katen-ka-za [31]2 years ago

6 0

Answer:

youtg jugre contact coming call details

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A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 37.0 vibrations

Goryan [66]

Answer:

0.47 m

Explanation:

$N$ = Number of vibrations = 37

$t$ = total time taken = 33 s

$T$ = time period of each vibration

frequency of vibration is given as

$f = \frac{N}{t} \\f = \frac{37}{33} \\f = 1.12$ Hz

$d$ = distance traveled along the rope = 421 cm = 4.21 m

$t$ = time taken to travel the distance = 8 s

$v$ = speed of the wave

Speed of the wave is given as

$v = \frac{d}{t}\\v = \frac{4.21}{8}\\v = 0.53 ms^{-1}$

$\lambda$ = wavelength of the harmonic wave

wavelength of the harmonic wave is given as

$\lambda = \frac{v}{f} \\\lambda = \frac{0.53}{1.12} \\\lambda = 0.47 m$

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Which element could be described by the list of properties above?

Murljashka [212]

Answer:

list the properties

Explanation:

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olga55 [171]

Answer:nah u took my points I take urs

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3 years ago

Which location on the map above is a source of North Atlantic deep water?

bija089 [108]

The answer might be C ? hope it's right

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3 years ago

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

Gnesinka [82]

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>
$\left | \overrightarrow{R} \right |=
\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}$

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

6 0

3 years ago