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3 years ago

Which kind of wave interaction is shown?

Physics

1 answer:

nlexa [21]3 years ago

3 0

Answer:

where?

Explanation:

ill edit my answer when see it

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Gravity does not actually "pull" objects at

Radda [10]

I believe it’s (D. Any object)

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3 years ago

The distance of Saturn from the sun is:<br><br> < 1 A.U.<br> > 1 A.U.<br> = 1 A.U.

sertanlavr [38]

Answer:

1 astronomical unit is the average distance from the Earth to the Sun; approximately 150 million km. At its closest point, Saturn is 9 AU, and then at its most distant point, it's 10.1 AU. Saturn's average distance from the Sun is 9.6 AU. We have written many articles about Saturn for Universe Today.

Explanation:

6 0

3 years ago

Read 2 more answers

A red cross helicopter takes off from headquarters and flies 120 km at 70 degrees south of west. There it drops off some relief

fiasKO [112]

Answer:

130 km at 35.38 degrees north of east

Explanation:

Suppose the HQ is at the origin (x = 0, y = 0)

So the coordinates of the helicopter after the 1st flight is

$x_1 = -120cos70^o = -41.04 km$

$y_1 = -120sin70^o = -112.763 km$

After the 2nd flight its coordinate would be:

$x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km$

$y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km$

So in order to fly back to its HQ it must fly a distance and direction of

$s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km$

$tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71$

$\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o$ north of east

3 0

3 years ago

Geologists have evidence that the continents were once a single giant landmass that land mass eventually split apart and the ind

mars1129 [50]

The part that it played was in the changing of the environment/climate for the organisms that live on those continents. And the part it could've played was the way that the organisms had to adapt to that climate and it stayed that way over generations.

7 0

3 years ago

A circuit contains a 305 ohm resistor, a 1.1 micro Farad capacitor, and a 42 mH inductor. At resonance, the impedance is determi

r-ruslan [8.4K]

Answer:

The resistance of the inductor at resonance is 258.76 ohms.

Explanation:

Given;

resistance of the resistor, R = 305 ohm

capacitance of the capacitor, C = 1.1 μF = 1.1 x 10⁻⁶ F

inductance of the inductor, L = 42 mH = 42 x 10⁻³ H = 0.042 H

At resonance the inductive reactance is equal to capacitive reactance.

$\omega L = \frac{1}{\omega C}\\\\2\pi F_0 L = \frac{1}{2\pi F_0 C}\\\\F_0 = \frac{1}{2\pi\sqrt{LC} }$

Where;

F₀ is the resonance frequency

$F_0 = \frac{1}{2\pi\sqrt{LC} } \\\\F_0 = \frac{1}{2\pi\sqrt{(0.024)(1.1*10^{-6})} }\\\\F_0 =980.4 \ Hz$

The inductive reactance is given by;

$X_l = 2\pi F_0 L\\\\X_l = 2\pi (980.4)(0.042) \\\\X_l = 258.76 \ ohms$

Therefore, the resistance of the inductor at resonance is 258.76 ohms.

5 0

2 years ago