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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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Does magnets exert a force. explain​
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Answer:

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3 years ago
What Is The frequency of an X Ray That had A Wavelength of 1.5* 10^-9
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Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
mylen [45]

Answer: 3.7 \times 10^{-4} N

Explanation:

The gravitational pull between two object is given by:

F = G\frac{Mm}{r^2}

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N

4 0
3 years ago
Read 2 more answers
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garri49 [273]
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3 years ago
According to Freud, adults fixated in which stage could feel constantly out of control?
yarga [219]

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