List three applications of pressure in our daily lives.​

All millisecond pulsars are now, or once were, members of binary-star systems. This statement is option a. true

Millisecond pulsars are now, or once were, members of binary-star systems. They are highly-magnetized, fast-spinning neutron stars formed through the transfer of mass from a companion star in a binary system. This transfer of mass leads to an increase in the pulsar's rotation speed, resulting in very rapid rotation periods, typically in the range of milliseconds.

Highly magnetised neutron stars known as pulsars produce electromagnetic radiation beams along their magnetic axes as they rotate. They were initially believed to be artificial signals from extraterrestrial intelligence when they were first found in 1967 by Jocelyn Bell Burnell and Antony Hewish. Pulsars can rotate up to several hundred times per second and release radiation at a variety of wavelengths, including radio waves and gamma rays. A number of astrophysical phenomena, including the behaviour of matter under extreme circumstances, the workings of gravity, and the makeup of the Milky Way galaxy, can be studied using the radiation emitted by pulsars.

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Answer:

like this

Explanation:

The correct answer is: All of these answers are correct.

All the provided explanations can be plausible reasons for the low adoption of a cookstove intervention in rural Bangladesh. It is important to consider factors such as technology maintenance, regional cooking preferences, and the suitability of the chosen stove for the user's level of education.

All of these factors can significantly impact the acceptance and adoption of a new cooking technology.It would hinder their ability to effectively use and benefit from the intervention.

Considering these factors collectively provides insight into why the adoption rate remained low. Addressing these issues is crucial to improving the acceptance and success of cookstove interventions in rural communities.

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Among the given options, cultural differences are most likely to be the centrifugal force within the EU in the future. The EU is a political and economic union of 27 member states, and cultural differences among its member states have always been present.

With the increasing number of immigrants from different parts of the world, the cultural differences among EU member states are becoming more prominent. Each member state has its unique language, history, customs, and traditions, which can create misunderstandings and conflicts among the member states.
The EU aims to promote unity and solidarity among its member states, but cultural differences can lead to a lack of understanding and trust between them. The EU's diverse cultural heritage is both a strength and a challenge for the union. The EU needs to find a way to respect the cultural diversity of its member states while maintaining its unity.
However, the cultural differences among the member states can still cause tensions and conflicts in the future. Therefore, it is crucial for the EU to continue to foster cultural awareness and understanding among its member states to maintain the union's cohesion and stability.

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complete question:

Which of these is most likely to be a centrifugal force within the EU in the future?

a. trade barriers

b. closed borders

c. pollution problems

d. cultural differences

The general equation of a circle in Cartesian coordinates is given by:

\((x - h)^2 + (y - k)^2 = r^2,\)

where (h, k) represents the coordinates of the center of the circle, and r represents the radius. Without the specific equation, we cannot determine if x and y satisfy it.

A circle is a two-dimensional geometric shape that is perfectly round and symmetrical. It is defined as a set of points that are equidistant from a central point called the center. The distance from the center to any point on the circle is called the radius, and it is the same for all points on the circle.

A circle is often represented by the symbol "⚪" or by writing its name. It is a fundamental concept in geometry and mathematics, and it has numerous properties and applications in various fields.

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The number of significant figures in the calculation for the perimeter should match the least number of significant figures among the measurements, which is 2. Therefore, the calculation for the perimeter should be reported with 2 significant figures.

When performing calculations, it is important to consider the rules of significant figures. The rule states that the result of a calculation should be rounded to match the least number of significant figures in the given measurements.

In the given measurements, the least number of significant figures is 2, which is found in Side 3 (2.000 m). Therefore, when calculating the perimeter of the seven-sided figure, the result should be rounded to 2 significant figures.

For example, if the calculated perimeter is 41.756 meters, it should be rounded to 42 meters since 2 is the least number of significant figures. This ensures that the result aligns with the precision of the measurements provided.

By adhering to the rules of significant figures, we can maintain consistency and accuracy in our calculations and ensure that the level of precision is appropriate for the given data.

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Answer:

falling 49m a second

Explanation:

Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

   PE = (2.5)(98)(14) and

   PE = 340 J

b. \(KE=\frac{1}{2}mv^2\) so

   \(KE=\frac{1}{2}(2.5)(14)^2\) and

   KE = 250 J

c. TE = KE + PE so

   TE = 340 + 250 and

   TE = 590 J

d. PE at 8.7 m:

   PE = (2.5)(9.8)(8.7) and

   PE = 210 J

e. The KE at the same height:

   TE = KE + PE and

   590 = KE + 210 so

   KE = 380 J

f. The velocity at that height:

   \(380=\frac{1}{2}(2.5)v^2\) and

   \(v=\sqrt{\frac{2(380)}{2.5} }\) so

   v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

   590 = KE + PE and

   PE = (2.5)(9.8)(11.6) so

   PE = 280 then

   590 = KE + 280 so

   KE = 310 then

   \(310=\frac{1}{2}(2.5)v^2\) and

   \(v=\sqrt{\frac{2(310)}{2.5} }\) so

   v = 16 m/s

h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

   \(v=v_0+at\) and

   26 = 0 + 9.8t and

   26 = 9.8t so the time at 26 m/s is

   t = 2.7 seconds. Now we use that in the equation for displacement:

   Δx = \(v_0t+\frac{1}{2}at^2\) and filling in the time the object was at 26 m/s:

   Δx = 0t + \(\frac{1}{2}(-9.8)2.7)^2\) so

   Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

Answer:

41.52 m

Explanation:

Using,

v = d/t....................... Equation 1

v = speed of pulse of sound, d = distance travelled by the pulse of sound, t = total time taken.

From equation 1, make d the subject of the equation

d = vt............................ Equation 2

Given: t = 0.12 s.

Constant: v = 346 m/s

Susbtitute these values into equation 2

d = 346(0.12)

d = 41.52 m

When a double-acting cylinder is operating at a single pressure, the force of extension will be equal to the force of retraction. This is because the cylinder has two ports that allow for pressure to be applied on both sides of the piston. As a result, the same amount of force is applied to extend and retract the piston.

However, there are certain factors that can affect the force of extension and retraction in a double-acting cylinder. These factors include the size of the piston, the pressure of the fluid, and the amount of friction in the cylinder. In some cases, the force of extension may be slightly greater than the force of retraction due to these factors.

It is important to note that the force of extension and retraction in a double-acting cylinder can be controlled by adjusting the pressure and flow of the fluid. This allows for precise control over the movement and force of the piston, making it an ideal choice for many applications in industries such as manufacturing, construction, and transportation.

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Larry Page, real name Lawrence Edward Page, is a computer scientist and entrepreneur from the United States who co-founded Goo-gle, one of the most well-known websites on the Internet, with Sergey Brin on March 26, 1973 in East Lansing, Michigan.

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Answer:

1) 2,641,895.5 m

2) 1.85903476 × 10⁻⁹ kg

3) 9.32 C

4) 38.7 kg

Explanation:

1) The gravitational force of attraction between the object and the mass of the Earth is given by the following formula;

\(F} =G\dfrac{M_{1} \cdot m_{2}}{R^{2}} = m_2 \times g\)

Where;

M₁ = The mass of the Earth = 5.97 × 10²⁴

m₂ = The mass of the object

G = The universal gravitational constant

R = The radius of the Earth = 6,378,100 m

g = The acceleration due to gravity = 9.8 m/s²

By comparison, we have;

\(\dfrac{G \cdot M_{1}}{R^{2}} = g\)

When the object is r meters away, g is halved, therefore, we have;

\(\dfrac{G \cdot M_{1}}{r^{2}} = \dfrac{g}{2} =\dfrac{G \cdot M_{1}}{\left (2\times R^{2}\right )}\)

Therefore, r² = 2·R² = 2 × (6,378,100 m)² = 8.13603192 × 10¹³ m²

r = √(8.13603192 × 10¹³ m²) = 9019995.52217 m

The distance of the object from the surface of the Earth = 9,019,995.52217 - 6,378,100 = 2,641,895.52217 m ≈ 2,641,895.5 m

The object needs to be approximately 2,641,895.5 m from the Earth's surface to experience half of the acceleration due to gravity experienced on the Earth's surface

2) The electrostatic force between the charges = k×q²/r²

Where;

q = 1.602 × 10⁻¹⁹ C

r = Distance between the charges

k = Coulomb constant = 8.9875517923 × 10⁹ kg·m³·s⁻²·C⁻²

We have;

The electrostatic force between the charges = 8.9875517923 × 10⁹ × (1.602 × 10⁻¹⁹)²/r²

The gravitational force between the charges = 6.67408 × 10⁻¹¹ × m²/r²

Given that the ratio between the two forces = 1, we have;

(8.9875517923 × 10⁹ × (1.602 × 10⁻¹⁹)²/r²)/(6.67408 × 10⁻¹¹ × m²/r²) = 1

∴ m² = (8.9875517923 × 10⁹ × (1.602 × 10⁻¹⁹)²)/(6.67408 × 10⁻¹¹) = 3.4560102×10⁻¹⁸ kg²

The mass m = √(3.4560102×10⁻¹⁸ kg²) = 1.85903476 × 10⁻⁹ kg

The mass of each particle will be 1.85903476 × 10⁻⁹ kg

3)  F = k×q²/r²

q² = F × r²/k

Which gives;

q² = 50 × 125000²/(8.9875517923 × 10⁹) ≈ 86.93 C²

q = √86.93 ≈ 9.32 C

The charges of each object = 9.32 C

4) \(F} =G\dfrac{m^2 }{r^{2}}\)

r = 0.25 cm = 0.0025 m

F = 0.016 N

G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²

Therefore;

m² = F·r²/G = 0.016 × (0.0025)²/(6.67408 × 10⁻¹¹) ≈ 1498.334 kg²

m = √(1498.334 kg²) ≈ 38.7 kg

The mass of each object = 38.7 kg.

The Coulomb and gravitational laws together with Newton's second law allow to find the results for the questions are:

    1) The distance where the acceleration of gravity measures half its value on earth is;  r’= 2.64 106 m

    2) The mass of the particle is:   m = 1.86 10-9 kg

    3) The charge of the particles is:   q = 9.32 C

    4) The mass is:   m = 38.7 kg

Part 1

The Universal Gravitational Law says that the force between two objects is proportional to their masses and inversely proportional to the square of their distance

     F =\(-G \frac{Mm}{r^2}\)  

Where F is the gravitational force, G the gravitational constant, M and m the mass of the two bodies and r the distance between them.

They ask to calculate the distance to the point where the acceleration is half that of the gravitation acceleration at the earth's surface.

            a = g / 2

If we use Newton's second law that gives the relationship between the net force, the mass and the acceleration of the body.

          F = m a

Let's substitute

        \(-G \frac{Mm}{r^2} = m \frac{g}{2}\)  

         r² = \(\frac{2GM}{g}\)  

Let's calculate

        r² = \(\frac{2 \ 6.67 \ 10^{-11} \ 5.97 \ 10^{24} }{9.8}\)  

        r = \(\sqrt{81.360 \ 10^{12}}\)  

        r = 9.02 10⁶ m

This is the distance from the center of the earth, the distance measured from the surface of the planet is

       r = R + r '

       r ’= r- R

       r ’= 9.02 10⁶ - 6.378 10⁶

       r’= 2.64 10⁶ m

Part 2

Coulomb's law states that the electric force between two charged bodies is:  

      F = \(k \frac{q_1q_2}{r^2}\)

Where q₁ and q₂ are the charges of the leather and k is the Coulomb constant

They ask what mass the particles must have so that the electrostatic force (Fe) and the gravitational force (Fg) have been equal

Indicate the charges of the particles q₁ = q₂ = 1.602 10⁻¹⁹ C

Let's equalize the two forces

     \(F_e = F_g \\k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{ r^2 } \\k q^2 = G m^2\)

     \(m = \sqrt{\frac{k}{G} } \ q\)  

     m = \(\sqrt{\frac{8.99 \ 10^9 }{ 36.67 \ 10^{-11}} } \ 1.602 \ 10^{-19}\)

     m =  \(\sqrt{1.3478 \ 10^{20} } \ 1.602 \ 10^{-19}\)  

    m = 1.86 10⁻⁹ kg

Part 3

They indicate that the electrostatic force between two objects is F = 50 N and the distance between them is r = 125 km = 125 103 m, they ask what is the charge of the objects.

Let's use Coulomb's law

        F = \(k \frac{q_1q_2}{r^2}\)  

        q² = \(\frac{F}{k} \ r^2\)  

        q² = \(\frac{50 \ (125 \ 10^3)^2 }{8.99 \ 10^9 }\)  

        q = \(\sqrt{86.93}\)  

        q = 9.32 C

Part 4

Indicate the gravitational force between two objects is F_g = -0.016 N when they are separateda distance  r = 0.25 cm = 0.25 10⁺² m, ask the mass of the bodies.

Let's use the law of universal gravitation.

      \(-G \frac{Mm}{r^2} \\m^2 = - \frac{F}{G} \ r^2\)

Let's calculate

      m² = \(- \frac{(-0.016) }{6.67 \ 10^{-11}} \ (0.25 \ 10^{-2})^2\)

     m = \(\sqrt{1498.33}\)  

     m = 38.7 kg

In conclusion using the Coulomb and gravitational laws together with Newton's second law we can find the results for the questions are:

   1) The distance where the acceleration of gravity measures half its value on earth is;  r’= 2.64 106 m

    2) The mass of the particle is:   m = 1.86 10-9 kg

    3) The charge of the particles is:   q = 9.32 C

    4) The mass is:   m = 38.7 kg

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Answer:

7/3 = 2 1/3

Explanation:

If the numerator is greater than or equal to the denominator of a fraction, then it is called an improper fraction. In that case, you could convert it into a whole number or mixed number fraction. 7/3 = 2 1/3.

Answer:

2.3333 (3-repeating) hope that helps

Answer:

ambt kaon✌️jhrhriekeieieieiei

Explanation:

iwuejejwiowksj

Kinetic energy = 29.912 J

Given

mass = 0.155 kg

vo=initial velocity = 31.4 m/s

h/d=30 m

Required

KE=kinetic energy

Solution

vt²=vo²-2ad⇒vt=final velocity

vt²=31.4²-2.10.30⇒g=10 m/s²

vt²=985.96-600

vt²=385.96 m/s

Kinetic  energy (KE)

\(\tt KE=\dfrac{1}{2}mv^2\\\\KE=\dfrac{1}{2}\times 0.155.\times 385.96\\\\KE=29.912~J\)

Answer:

The answer is 30.8 (J)

Explanation:

You're welcome

Answer:

Explanation: The density of gas differentiates all the varieties of gas present in the atmosphere.

HYDROGEN- it is a lighter gas having a lesser density.

For example, a hydrogen balloon displaces the air around it thus making it rise.

CARBON DIOXIDE: It is a denser gas as it makes the balloon heavier than the air pressure around it thus making it sink.

I hope the explanation was helpful. :)

Hi there!

When finding work (J) given force (N), we use the equation:

W = FdcosФ <-- Only take into account the force in the direction of motion

Plug in the given values:

W = 27(32)cos(20) = 811.89 J

Explanation:

Here,

Given that,

pulls a wagon a distance of 32 meters the handle is above an angle of 20 degrees above horizontal if you pull on it with a force of 27 N

To find,

As we know that,

\({\boxed{\sf{W=f.d.cos\theta }}}\)

According to the question,

\(W=27(32)cos{20}\\\\w=811.89J\)

The correct order for the life cycle of a high mass star are nebula, protostar, main sequence star, red supergiant, supernova, neutron star or black hole. The correct answer is C.

The life cycle of a high mass star begins with a cloud of gas and dust known as a nebula. Gravitational forces cause the nebula to collapse, forming a protostar, which grows in size as it continues to accrete matter.

Once the temperature and pressure at the core of the protostar are high enough, nuclear fusion reactions begin, and the protostar becomes a main sequence star.

As the star burns through its hydrogen fuel, it expands and becomes a red giant. Eventually, the core contracts and heats up, causing the outer layers to be expelled in a planetary nebula, leaving behind a hot, dense core known as a white dwarf.

If the star is massive enough, the core will continue to contract until it reaches a critical density, at which point it will collapse and explode in a supernova. The remnant of the supernova can either be a neutron star or a black hole, depending on the mass of the original star.

So, the correct order for the life cycle of a high mass star is: nebula, protostar, main sequence star, red supergiant, supernova, neutron star or black hole. Therefore, option C is the correct answer.

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Answer:

B. It's nucleus releases energy

Explanation:

Answer:

C. 1.42 s

Explanation:

Givens:

length = 0.5m

gravity = 9.807 m/s

Solve:

T = 2π√(L/g)

T = 2π√(0.5/9.807)

T = 2π√(0.0509)

T = 2π * 0.22579

T = 6.28318 * 0.22579

T = 1.418 ≈ 1.42s

Placing the electromagnetic radiation in order from the lowest energy to the highest energy :  ( 2 ) C,A,B

In the electromagnetic spectrum the electromagnetic radiation with the shorter wavelength possess a higher energy while the electromagnetic radiation with a longer wavelength possess the lower energy.

The electromagnetic radiation as listed in the question with the longest wavelength is the radio waves therefore it possess the lowest energy while the radiation with the shortest wavelength is the gamma rays therefore it possess the highest energy.

Hence we can conclude that Placing the electromagnetic radiation in order from the lowest energy to the highest energy :  ( 2 ) C,A,B

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Answer:

b. between 2 s and 4 s

Explanation:

2-4 was both 3m/s

Within the snow line planets are, Mercury, Venus, Earth, and Mars were all formed.

As the solar system was forming, a line existed between the orbits of Mars and Jupiter; as a result, the rocky planets Mercury, Venus, Earth, and Mars formed within the line, while Jupiter, Saturn, Uranus, and Neptune formed outside.

The frost line, which is sometimes referred to as the snow line or ice line, is the precise location in the solar nebula away from the central protostar where it is cold enough for vapors like water, ammonia, methane, carbon dioxide, and carbon monoxide to condense into solid ice grains.

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Answer:The velocity of the object will be 5\(\sqrt7\)m/s or 13.23m/s

Explanation:

force exerted by the object= 30N

distance displayed by the object by the action of force=6.0m

mass of object=10kg

velocity gained by the object=?

\(\frac{1}{2}mv^{2}= forcexdisplacement\\\frac{1}{2}10v^{2} = 30x6\\ 5v^{2}=180\\ v^{2}= 180-5\\ v^{2} =175\\v=\sqrt{175} \\v=5\sqrt{7} or 13.23\)

Answer:

As the temperature is increased the more the energy store

Answer:

\( \huge \boxed{ \boxed{2 + \sqrt{3}}} \)

Explanation:

2. The primary reflection from the sand/mud interface will arrive first at the hydrophone. To determine which event arrives first, we need to calculate the two-way travel times (TWTT) for each event. The TWTT for the primary reflection from the sand/mud interface is:

TWTT = (2 × depth × sin (angle of incidence)) / velocity

TWTT = (2 × 509 × sin (9)) / 1478TWTT = 0.317 s

The TWTT for the double bounce off the seafloor is:TWTT = (2 × depth) / velocityTWTT = (2 × 509) / 1478TWTT = 0.689 s

Therefore, the primary reflection arrives first at the hydrophone. Here is a simple drawing of the two-event seismic trace:

3. To calculate the time it takes for energy to travel directly from the air gun to the first hydrophone, we need to determine the distance between them and divide it by the velocity of sound in seawater. Using the given values, we have:

Distance = depth + (thickness of sand/mud) + (thickness of mudstone)

Distance = 509 + 1003 + 373

Distance = 1885 m

Velocity of sound in seawater = 1478 m/s

Time = Distance / VelocityTime = 1885 / 1478Time = 1.276 s

Therefore, it takes 1.276 seconds for energy to travel directly from the air gun to the first hydrophone.

4. The maximum takeoff angle at which seismic energy can reflect from the sand/mud interface is called the critical angle. This angle can be calculated using Snell's law:

n1 × sin (angle of incidence) = n2 × sin (angle of refraction)

where n1 and n2 are the velocities of the two materials and the angle of refraction is 90 degrees (since seismic energy travels along a horizontal path once it reaches the interface).

For the sand/mud interface, the critical angle is:

n1 × sin (critical angle) = n2 × sin (90)n1 / n2 = cos (critical angle)critical angle = cos^-1 (n1 / n2)

Using the given values:

n1 = 2793 m/s (sandstone velocity)n2 = 2240 m/s (mudstone velocity)critical angle = cos^-1 (2793 / 2240)

critical angle = 35.9 degrees

Seismic energy cannot reflect from the sand/mud interface at angles greater than the critical angle. For larger angles, the energy will be transmitted into the mudstone.

5. When seismic energy is transmitted into the mudstone, it travels in all directions away from the source. However, the energy will be attenuated (reduced in amplitude) as it travels through the mudstone due to its relatively low velocity compared to the sandstone and seawater.

As a result, the mudstone acts as a barrier that blocks or reduces the energy that would otherwise be transmitted deeper into the subsurface.

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Answer:

Explanation:

The mass of the arrow can be found by using the formula

\(m = \frac{p}{v} \\ \)

p is the momentum

v is the velocity

From the question we have

\(m = \frac{5.22}{48.3} \\ = 0.108074...\)

We have the final answer as

Hope this helps you

The period for a small amplitude oscillation on a thin, straight, uniform rod 1.10 m long and 250 kg mass hangs from a pivot at one end = 2.09 s.

Harmonic vibration is the alternating motion of an object through an equilibrium point that has a fixed frequency and period. An example of harmonic vibration is a pendulum.

Period (T) is the time needed to do one vibration. Meanwhile, the frequency (f) is the number of vibrations that occur in one second. The period and frequency of the pendulum depend on the length of the string and the acceleration due to gravity, and are independent of the mass of the pendulum. The period and frequency of the pendulum can be expressed using the following equation:

T  = 2\(\pi\)\(\sqrt{\frac{L}{g} }\)

f = \(\frac{1}{2\pi }\)  \(\sqrt{\frac{L}{g} }\)

Where,

T = period (s)

f = frequency (Hz)

L = length (m)

g = acceleration of gravity (9.8 m/s²)

We have,

L = 1.10 m

So,

T = 2\(\pi\)\(\sqrt{\frac{L}{g} }\)

= 2\(\pi\)\(\sqrt{\frac{1.10}{9.8} }\)

= 2.09 s

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Answer:

Explanation:

where is the question