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3 years ago

According to Boyle’s Law, when pressure is increased...

1 answer:

4 0

Boyle's law states that when you "shrink"(change its size) a container the pressure increases so the best answer would be D-volume is decreased.

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Fast!! Help me!!<br>Subject- Science

Archy [21]

Last One.... If Im not Wrong!

Burning of Fossil Fuels, Increase the CO2 And therefore the Green Effect!!

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3 years ago

Paolo says that his bicycle is hard to pedal.Mia looks at the bicycle and tells him he needs to oil the chain.explain why oiling

laila [671]

Answer:

Less friction

Explanation:

Paolo's bike is too difficult to pedal because there is too much friction in the mechanisms of the bike. To reduce friction, Paolo must oil the chain. This will make the bike run much more smoothly and allow for easier pedalling.

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2 years ago

Select the correct answer.

Liono4ka [1.6K]

Answer: The electromagnetic waves reach Earth, while the mechanical waves do not.

Explanation:

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3 years ago

Read 2 more answers

Compare the gravitational acceleration on the following objects compared to the Sun using:

arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0

2 years ago

At about 55 meters/sec, a falling parachuter (before the parachute opens) no longer accelerates. Air friction opposes accelerati

JulijaS [17]

The acceleration of gravity on Earth is about 9.8 m/s².
That means that if air resistance is neglected, a falling body
smoothly gains 9.8 m/s of downward speed each second.

If our intrepid chutist lightly stepped out of the plane or was
gently pushed, so that his initial downward speed was zero
and grew by 9.8 m/s every second, then it took him ...

(55) / (9.8) = 5.6 seconds

to reach that particular downward speed.

3 0

3 years ago