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Katarina [22]
3 years ago
6

Which two technologies use the same range of radio wave frequencies? television broadcasting and AM radio broadcasting cell phon

es and space communication CB radios and submarine communication AM and FM radio broadcasting
Physics
2 answers:
jolli1 [7]3 years ago
8 0

Answer:

The correct answer is cell phones and space communication.

Explanation:

I took the test on edge.

Tju [1.3M]3 years ago
4 0
Let us reject some answers. AM and FM radio waves do not interfere. Am are an orders of magnitude larger than FM waves. FM waves belong to microwaves, a subcategory of radiowaves while AM does not. Similarly, AM  and TV have not the same, because TV uses a subset of the frequencies used in FM. The correct answe is that CB radios and submarine communication happen at the same wavelengths/frequencies, namely low frequency radiowaves.
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A switch can
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two masses are separated by 1 m. suppose the masses are moved so they are 2 m apart how will the gravitational force change
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The gravitational force would get stronger because the farther the two masses are separated the more gravitational force will be used to pull them together the closer they are the less gravitational pull is used to pull them together
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What distance is covered by an airplane traveling at a velocity of 660 miles per hour in 3.5 hours?
N76 [4]

As per the question, the velocity of the airplane [v] = 660 miles per hour.

The total time taken by airplane [t] = 3.5 hours.

We are asked to determine the total distance travelled by the airplane during that period.

The distance covered [ S] by a body is the product of velocity with the time.

Mathematically distance covered = velocity × total time

                                                      S = v × t

                                                        = 660 miles/hour ×3.5 hours

                                                        = 2310 miles.

Hence, the total distance travelled by the airplane in 3.5 hour is 2310 miles.

4 0
3 years ago
A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein
RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 8m^{3}/s. As you  may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

V_{cone}=\frac{1}{3} \pi r^{2}h

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

\frac {r}{h}=\frac{4}{5}

When solving for r, we get:

r=\frac{4}{5}h

so we can substitute this into our volume of a cone formula:

V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h

which simplifies to:

V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h

V_{cone}=\frac{16}{75} \pi h^{3}

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}

Which simplifies to:

\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}

Now we can substitute the provided values into our equation. So we get:

\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}

so:

\frac{dh}{dt}=0.25m/s

3 0
3 years ago
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