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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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The total mass of the train and its passengers is 750000kg. The train is traveling at a speed of 84m/s. The driver applies the b
fiasKO [112]

Answer:

|F| = 393750  N

Explanation:

Given that,

Total mass of the train, m = 750000 kg

Initial speed, u = 84 m/s

Final speed, v = 42 m/s

Time, t = 80 s

We need to find the net force acting on the train. The formula for force is given by :

F = ma

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{750000\times (42-84)}{80}\\\\F=-393750\ N

So, the magnitude of net force is 393750  N.

4 0
3 years ago
What is matter? Give a grade 10/11/12 understanding.
lara [203]

Matter can be many things, but is mostly made of atoms. Atoms are small things that cannot be seen directly, as lights either passes through it or alters it. However, we know that atoms are made of 3 parts, those being the proton, neutron, and electron.

Protons have a positive electromagnetic charge.

Neutrons have no charge.

Electrons have a negative charge equal to the protons positive charge.

Protons and neutrons make up what's called the nucleus, which is orbited by the electrons.

Protons and neutrons also share another thing in common, that being their composition.

Until relatively recently, we thought that these were the smallest particles in the universe, and indestructible. However, modern discoveries have revealed that they are actually made of quarks and gluons.

These are actually indestructible, being part of the group that is elementary particles.

3 0
3 years ago
When objects are forced to vibrate ( like when dropped onto a hard surface), they will do so at their
Marat540 [252]

They'll vibrate at their characteristic resonant frequency. That depends on the material the object is made of and its shape.


3 0
3 years ago
A ball is thrown upward at a 45° angle. Inthe absence of air resistance, the ballfollows aA. tangential curve.B. sine curve.C. p
Evgesh-ka [11]

As ball is projected up in air at an angle of 45 degree without any air resistance

Let the initial speed will be v

now we will have

In x direction

v_x = v cos45

in y direction

v_y = vsin45

now displacement in x direction

x = (vcos45)t + 0

displacement in y direction

y = (vsin45)t - \frac{1}{2}gt^2

now from above two equations we have

y = (vsin45)\frac{x}{vcos45} - \frac{1}{2}g(\frac{x}{vcos45})^2

y = xtan45 - \frac{1}{2v^2cos^245}gx^2

so above equation is a quadratic equation and hence it will be a parabolic curve

so correct answer will be

<em>C. parabolic curve.</em>

8 0
3 years ago
For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
3 years ago
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