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2 years ago

If Spacecraft X has twice the mass of Spacecraft Y , then what is true about X and Y ?

Physics

1 answer:

Solnce55 [7]2 years ago

6 0

I don't know what you were trying to copy or type in Choice-E .

All three statements are true, so the answer should say "I, II, and III".

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The diagram shows a skydiver at different points of her jump. At what point would her

8_murik_8 [283]

I’m pretty sure it’s B! :)

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2 years ago

State coulombs law in word

amid [387]

<h2>state coulombs law in word</h2>

: a statement in physics: the force of attraction or repulsion acting along a straight line between two electric charges is directly proportional to the product of the charges and inversely to the square of the distance between them

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2 years ago

Four importance of soil water

Andru [333]

Answer:

nitrogen, phosphorus, potassium, and calcium

Explanation:

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2 years ago

Meteorology is BEST defined as

alexgriva [62]

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2 years ago

Read 2 more answers

A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp

NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

$f=12GHz=12\cdot 10^9 Hz$ is the frequency

$c=3.0\cdot 10^8 m/s$ is the speed of light

So the wavelength of the beam is

$\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m$

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

$y=\frac{m\lambda D}{a}$

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

$y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m$

and so, the diameter is

$d=2y = 750 m$

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

$r=\frac{750 m}{2}=375 m$

So the area is

$A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2$

And since the power is

$P=100 kW = 1\cdot 10^5 W$

The average intensity is

$I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2$

4 0

2 years ago