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2 years ago

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1 answer:

4 0

According to Archimede's principle, a physical object experiences an upthrust due to a difference in pressure between upper and lower surfaces.

<h3>What is an upthrust?</h3>

An upthrust is also referred to as buoyancy and it can be defined as an upward force which is exerted by a fluid (liquid or gas), so as to oppose the weight of a partially or fully immersed physical object that is floating in it.

Based on scientific information, a physical object experiences an upthrust when it is immersed in a fluid due to a difference in height and pressure between upper (top) and lower (bottom) surfaces.

According to Archimede's principle, there is a higher pressure at the bottom of the physical object due to height, and a lower pressure at the top of the physical object.

Read more on upthrust here: brainly.com/question/24389514

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Which planet search technique is currently best suited to finding earth-like planets?

Elis [28]

Answer:

The planet search technique currently best suited to find Earth-like planets is gravitational microlensing.

Explanation:

<u><em>Gravitational microlensing</em></u>

When a foreground star (lensing star) passes in front of a background star (source star), due to its gravitational field it warps space and magnifies the light coming from the source star. If the lensing star has a planet orbiting near, the planet's gravity also bends light and intensifies this effect. What is observed is a sharp increase in brightness in the otherwise regular pattern of the microlensing event. Certain characteristics of the planet like its total mass, orbit and period can be obtained from said pattern.

This technique is independent of the wavelength of the source star which makes it suitable for any kind of electromagnetic radiation.

Microlensing is suitable to find smaller, rocky planets like Earth because is more sensitive to planets whose orbits are further apart from the parent star.

Other indirect detection techniques like the radial velocity method and the transit method are biased towards massive gaseous planets that orbit very close to their parent star.

The <em>radial velocity method</em> makes use of the Doppler effect, that involves the change in frequency of a wave depending on the relative movement of the observer and the wave source. This relative motion, that should be in the line that joins the wave source and the observer, is called the radial motion. That is why the velocity of this motion is called the radial velocity. If a star is moving towards Earth the light waves reach us faster. We say the spectrum is blue shifted because the color blue has the shortest wavelength in the visible spectrum. If the star is moving away from the Earth the light waves reach us later and the wavelength becomes larger. We say the spectrum is red shifted because the color red has the longest wavelength. If a planet is orbiting star there will be stellar motion caused by the tug of the planet, the doppler shift allow us to detect this subtle motion. Is currently unsuitable to detect small, rocky planets like Earth because maasive planets orbiting very close to their stars create a larger and easily to note spectral shift.

<em>Transit method </em>: if a planet crosses in front of the star it orbits, the star brightness tenporarily decreases a little. This method is also unsuitable because with large and gaseous planets the drop in brightness iseasier to spot.

6 0

3 years ago

The solution to a disjunction is the what of the two solutions

astra-53 [7]

Answer:

The solution set of a disjunction is the union of the solution sets of the individual inequalities. A convenient way to graph a disjunction is to graph each individual inequality above the number line, then move them both onto the actual number line

Explanation:

6 0

3 years ago

What is necessary for a metallic bonds to form

Rus_ich [418]

Metallic bonds can occur between different elements to form alloy. In contrast to electrons that participate in both ionic and covalent bonds electrons are that participate in metallic bonds delocalize forming a sea of electrons around the positive nuclei of metals.

7 0

3 years ago

Read 2 more answers

A narrow beam of light containing orange (610 nm) and blue (470 nm) wavelengths goes from polystyrene to air, striking the surfa

Eva8 [605]

Answer: The angle between the two colours when they emerge is 0.4°

Explanation:

The refracting angle for different colors knowing its refractive index can be calculated using Snell's law.

Ni × sin α = Nr × sinβ

Where Ni is the refractive index for light in incident medium

α is the angle the incident makes with normal

Nr is the refractive index for light in the refractive medium

β is the angle the refracted makes with the normal

Making β the subject

β = arcSin ( Ni × sinα)/Nr

For Orange color

The value of refractive index in polystyrene medium is 1.490 ,that is Ni= 1.490

α is 30° and Nr is the refractive index of air = 1

β = arcSin(1.490×sin30°)/1

β= 48.15°

For Blue color

The value of refractive index in polystyrene medium is 1.499,that is Ni= 1.499

α is 30° and Nr is the refractive index of air = 1

β = arcSin(1.499×sin30°)/1

β= 48.55°

The angle between the two colours is the difference in the angles of their refracted rays

48.55-48.15=0.4°

6 0

3 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is: