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BaLLatris [955]
3 years ago
11

A neutral atom of an element has the same number of __________ and ________. Question 6 options: Neutrons and electrons Protons

and neutrons Protons and electrons Save
Physics
1 answer:
Bess [88]3 years ago
8 0
The answer is:  [C]:  "protons and electrons" .
____________________________________________________
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OLEGan [10]

Answer:

A. Yard

Explanation:

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Marine science- question- what is climate change?
Troyanec [42]

Answer:

Defined below

Explanation:

Climate change is simply the long term change in the average weather patterns that are associated with the local, regional and global climates of the earth. Climate change is usually driven by human activities like burning of fossils; natural processes like cyclic ocean patterns; external factors like volcanic eruptions.

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3 years ago
Are amplitude and wavelength related?<br> A: Yes<br> B: No
Trava [24]

The answer is A.Yes

Explanation:

The amplitude of a wave is the height of a wave as measured from the highest point of the wave to the lowest on the wave.

4 0
2 years ago
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
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Please help! Will mark best answer!!
MA_775_DIABLO [31]
The answer is (A) hope it helps 
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