Which of the following types of rocks could contain organic matter

what would happen to the temperature on mercury during the day if a sunspot was directly in line with it's surface?

QveST [7]

Answer:

mind-bogglingly hot

Explanation:

If a sunspot were pointed right at Mercury in this way, the planet would become even more mind-bogglingly hot. Sunspots look dark in our visual spectrum, so we might suspect that they are colder (or at least less hot) than the rest of the Sun; but in fact, they are hotter.

3 0

2 years ago

Which formula represents Charles’s law? P1V1 = P2V2 V1T1 = V2T2

Lynna [10]

Charles's law is best represent by the formula, V1T2= V2T1.

Charles's law state that for an ideal gas, at constant pressure, the absolute volume of a gas is directly proportional to the temperature.

that is

V α T

Or

V1T2= V2T1.

3 0

3 years ago

Read 2 more answers

How does the gravitational pull of the blue supergiants impact the direction of your star?

Gennadij [26K]

Answer:

Blue supergiants represent a slower burning phase in the death of a massive star. Due to core nuclear reactions being slightly slower, the star contracts and since very similar energy is coming from a much smaller area (photosphere) then the star's surface becomes much hotter.

Explanation:

I know this may not be the answer youre looking for, but hopefully this can help somehow!

8 0

2 years ago

5. The boiling point of mercury is lower than<br> that of alcohol.

aniked [119]

Answer:

I thinl it is I'm not sure though

3 0

3 years ago

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A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

(c) From any of the equations, solve for Length of the string (L);

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$

(a) the fundamental frequency is calculated as;

$f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz$

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

7 0

2 years ago