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2 years ago

Classify the following matter

Physics

1 answer:

AURORKA [14]2 years ago

7 0

Answer:

Woah

Explanation:

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How does energy acquisition in the deep sea differ from energy acquisition near the ocean’s surface? a. Organisms in the deep se

Montano1993 [528]

that would be the energy from the core. because of the fact that ht eearths core is producing 45mill killawhats and giggawats

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3 years ago

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NEED HELP!! ASAP JUST QUESTIONS 22, 23, 24, & 25

poizon [28]

22. reduction
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3 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:

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

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3 years ago

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about how much more energy is released in a 6.5 richter magnitude earthquake than in one with magnitude 5.5?

OverLord2011 [107]

Answer:

For example, an earthquake of magnitude 5.5 releases about 32 times as much energy as an earthquake measuring 4.5. Another way to look at this is that it takes about 900 magnitude 4.5 earthquakes to equal the energy released in a single 6.5 earthquake.

Explanation:

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2 years ago

The security alarm on a parked car goes off and produces a frequency of 769 Hz. The speed of sound is 343 m/s. As you drive towa

uysha [10]

Answer:

Explanation:

ASSUMING your speed is constant

f₀ = f(v + vo)/(v + vs)

Δf = f approach - f depart

69.5 = (769(343 + vo)/(343 + 0)) - (769(343 - vo)/(343 + 0))

69.5 = 769(2vo/343)

vo = 15.5 m/s

6 0

2 years ago