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avanturin [10]
3 years ago
12

PLEASE HELP, I am desperate. This part of the presentation I am completely lost with. It is STEP 6, I got the rest finished.

Physics
1 answer:
yKpoI14uk [10]3 years ago
5 0

Answer:

Explanation:

Ok, pick some idea from here

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=10&ved=2ahUKEwjWoLaFypvnAhUowlkKHXZrCMkQFjAJegQIBBAB&url=https%3A%2F%2Fwww.wappingersschools.org%2Fcms%2Flib%2FNY01001463%2FCentricity%2FDomain%2F1552%2FGravity.ppt&usg=AOvVaw1N8kqPWy6ERpA3LNbbepit

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An airplane flies 40km directly south in 10 minutes and 20 km directly east in five minutes, its average speed is:
blondinia [14]
Average speed = total distance / total time
total distance = 40 + 20 = 60km
total time taken = 10 + 5 = 15 minutes
Average speed = 60/15 = 4km/min
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How are ice liquid water and water vapor different from each other?
olga_2 [115]
The molecules of ice stick together in the process of cohesion. They are tightly packed so there isn't much room to move. Liquid water is a looser hold. The molecules can go past one another, and they will take the shape of whatever container they occupy. Water vapor is loosely contained, and it will will fill whatever container it is kept in, and it will take its shape, too.
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A procedure called______is used to find the volume of a rock
KengaRu [80]

Displacement is usually how the messure the rock

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A tall, open container is full of glycerine. At what depth h below the surface of the glycerine is the pressure 2370 Pa greater
Mice21 [21]

Answer:

So, at the depth of 24 cm below the surface of the glycerine the pressure is  2970 Pa. Hence, this is the required solution.

Explanation:

Given that,

Pressure exerted by the surface of glycerine, P = 2970 Pa and it is greater than atmospheric pressure.

The density of glycerine,  

We need to find the depth h below the surface of the glycerine. The pressure due to some depth is given by :

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3 years ago
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
3 years ago
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