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2 years ago

Does a closed system obey the law of conservation of matter? Write one or two sentences explaining your answer

1 answer:

5 0

Yes, a closed system obeys the law of conservation of matter. The law states that matter cannot be created not destroyed, only transferred. Meaning that in a closed system, this matter will be lost from one object but a different object will gain the same matter that was lost, therefore only transferring it. If it were an open system, the matter would be transferred outside the system and therefore you would loose matter on the system, but still, matter was not destroyed or created, it was only transferred from one thing to another.

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(03.05 LC)<br> Which of the following best describes the use of a renewable resource

RUDIKE [14]

Answer:

Option C or the third option.

Explanation: Water is a renewable resource there is so much of it and it just keeps circulating through the system it doesn't run out.

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3 years ago

A charge of 5.67 x 10^-18 C is placed 3.5 x 10^-6 m away from another charge of -

Fudgin [204]

+ 1.58 e -15

Please hit thanks button! :)

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2 years ago

Mali loves to make herself dizzy. She spins in place 7 times before falling down right where she was standing. Find her distance

Mila [183]

The answer is 0 bc she didint move
Explanation:

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3 years ago

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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

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

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3 years ago

Which feature or process is unique to nuclear power plants when compared to conventional coal-burning power plants?

Mazyrski [523]

Answer:

Control rods

Explanation:

Control rods is unique to nuclear power plants as they are mainly used in nuclear reactors to stabilize the rate of fission of plutonium or uranium. They are usually composed of chemical elements such as cadmium, silver, boron, and indium.

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3 years ago

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