A beta and gamma source, Co-60, was placed in a chamber which was first lined with paper and then, after, with copper and lead. A Geiger counter was used in both cases to detect the emission of radiation. You expect

Answer: trial 3

Explanation:

impulse equals force x time so for each trial it would be a force which is 500 x 8 individual time trial 3 has the highest time which would equal the highest impulse.

500 x .45 = 225

Answer:

Trial 3

Explanation:

got it correct

B is the correct answer

Hope this helps:)

The correct path of light through a reflecting telescope will be Primary mirror--->secondary mirror--->eyepiece--->eye.

A reflecting telescope (also called a reflector) is a telescope that uses a single or a combination of curved mirrors that reflect light and form an image

it is a design that allows for very large diameter objectives. Almost all of the major telescopes used in astronomy research are reflectors.

Reflecting telescopes come in many design variations and may employ extra optical elements to improve image quality or place the image in a mechanically advantageous position.

A curved primary mirror is the reflector telescope's basic optical element that creates an image at the focal plane. The distance from the mirror to the focal plane is called the focal length.

Film or a digital sensor may be located here to record the image, or a secondary mirror may be added to modify the optical characteristics and/or redirect the light to film, digital sensors, or an eyepiece for visual observation.

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Answer:

A virtual image is formed at the position from which the rays appear to have originated

Explanation:

When using lenses or mirrors to produce the image of an object, two types of images can be produced:

- Real image: a real image is produced when the rays of light actually converge to a point - in this case, the image can be projected on a screen. For a lens, a real image is located on the opposite side of the lens relative to the object, while for a mirror, a real image lies on the same side of the object with respect to the mirror

- Virtual image: a virtual image is formed at the position from which the rays appear to have originated. Because of that, a virtual image cannot physically be projected on a screen. For a mirror, a virtual image is located on the opposite side of the mirror relative to the object, while for a lens, a virtual image lies on the same side of the object with respect to the lens.

When a wave travels through a material, it can either reflect, refract or diffraction.

When a wave travels through a material, it interacts with the material in a number of ways, depending on the type of wave and the properties of the material. It could result into reflection, refraction and diffraction.

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Answer:

[tex]2.46\cdot 10^5 J[/tex]

Explanation:

The enegy of a single photon is given by:

[tex]E=\frac{hc}{\lambda}[/tex]

where

h is the Planck costant

c is the speed of light

[tex]\lambda[/tex] is the wavelength of the photon

In this problem,

[tex]\lambda=486 nm=4.86\cdot 10^{-7}m[/tex]

so the energy of one photon is

[tex]E_1=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.86\cdot 10^{-7}m}=4.09\cdot 10^{-19} J[/tex]

1 mole of photons contains a number of Avogadro of photons:

[tex]N_A = 6.022\cdot 10^{23}[/tex]

therefore, the total energy of 1 mole of these photons will be

[tex]E=N_A E_1 = (6.022\cdot 10^{23})(4.09\cdot 10^{-19} J)=2.46\cdot 10^5 J[/tex]

The energy of a mole of photons with a wavelength of 486 nm is 2.462 x 10^5 joules, calculated by applying Planck's equation and using Avogadro's number.

The energy in joules of a mole of photons associated with visible light of wavelength 486 nm can be calculated using Planck's equation, E = hc/\u03bb, where h is Planck's constant (6.626 x 10-34 J·s), c is the speed of light (3.00 x 108 m/s), and \u03bb is the wavelength of the light (486 nm or 486 x 10-9 m). To find the energy per mole of photons, we also use Avogadro's number (6.022 x 1023 photons/mole).

First, calculate the energy of one photon:

E = hc/\u03bb = (6.626 x 10-34 J·s) (3.00 x 108 m/s) / (486 x 10-9 m) = 4.09 x 10-19 J/photon

Then, calculate the energy per mole of photons:

Emole = E · Avogadro's number = 4.09 x 10-19 J/photon x 6.022 x 1023 photons/mole = 2.462 x 105 J/mole

Therefore, the energy of a mole of photons at 486 nm is 2.462 x 105 joules.

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When two charges of opposite sign are placed near each other, the electric potential energy decreases while the kinetic energy increases.

When two charges of opposite sign are placed near each other, the electric potential energy, kinetic energy, and work change in specific ways. Initially, the charges have potential energy due to their position in the electric field. As they move closer together, the potential energy decreases, and this decrease is converted into kinetic energy.

The work done on the charges is negative because energy is being taken away from the system. In other words, the charges are pulling on each other and you need to do work to bring them closer. Overall, the potential energy decreases, and the kinetic energy increases.

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When two charges of opposite sign are placed near each other, the electric potential energy decreases while the kinetic energy increases. Work is done as the charges are brought near each other.

When two charges of opposite sign are placed near each other, the electric potential energy changes. The potential energy decreases as the charges approach each other, resulting in a decrease in potential energy. As the potential energy decreases, the kinetic energy of the charges increases. This is because the electric field between the charges accelerates the charges, converting their potential energy into kinetic energy. Lastly, work is done when the charges are brought near each other, as the electrostatic force between the charges can do work on the charges.

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Answer:

B

Explanation:

Both low and high mass stars begin as nebulae, then become protostars.  Both use nuclear fusion to form hydrogen in the main sequence.

The differences are that low mass stars have longer life cycles and become white dwarfs.  High mass stars have shorter life cycles and undergo supernova explosions.

Answer: b

Explanation:

A) [tex]E(r) = \frac{\rho r_b^3}{3 \epsilon_0 r^2}[/tex]

In this problem we have spherical symmetry, so we can apply Gauss theorem to find the magnitude of the electric field:

[tex]\int E(r) \cdot dr = \frac{q}{\epsilon_0}[/tex]

where the term on the left is the flux of the electric field through the gaussian surface, and q is the charge contained in the surface.

Here we are analyzing the field at a distance [tex]r>r_B[/tex], so outside the solid ball. If we take a gaussian sphere with radius r, we can rewrite the equation above as:

[tex]E(r) \cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex] (1)

where [tex]4 \pi r^2[/tex] is the surface of the sphere.

The charge contained in the sphere, q, is equal to the charge density [tex]\rho[/tex] times the volume of the solid ball, [tex]\frac{4}{3}\pi r_b^3[/tex]:

[tex]q= \rho (\frac{4}{3}\pi r_b^3)[/tex] (2)

Combining (1) and (2), we find

[tex]E(r) \cdot 4 \pi r^2 = \frac{4\rho \pi r_b^3}{3 \epsilon_0}\\E(r) = \frac{\rho r_b^3}{3 \epsilon_0 r^2}[/tex]

And we see that the electric field strength is inversely proportional to the square of the distance, r.

B) [tex]\frac{\rho r}{3 \epsilon_0}[/tex]

Now we are inside the solid ball: [tex]r<r_B[/tex]. By taking a gaussian sphere with radius r, the Gauss theorem becomes

[tex]E(r) \cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex] (1)

But this time, the charge q is only the charge inside the gaussian sphere of radius r, so

[tex]q= \rho (\frac{4}{3}\pi r^3)[/tex] (2)

Combining (1) and (2), we find

[tex]E(r) \cdot 4 \pi r^2 = \frac{4\rho \pi r^3}{3 \epsilon_0}\\E(r) = \frac{\rho r}{3 \epsilon_0}[/tex]

And we see that this time the electric field strength is proportional to r.

C)

E(0)=0.

limr→∞E(r)=0.

The maximum electric field occurs when r=rb.

Explanation:

From part A) and B), we observed that

- The electric field inside the solid ball ([tex]r<r_B[/tex]) is

[tex]\frac{\rho r}{3 \epsilon_0}[/tex] (1)

so it increases linearly with r

- The electric field outside the solid ball ([tex]r>r_B[/tex]) is

[tex]E(r) = \frac{\rho r_b^3}{3 \epsilon_0 r^2}[/tex] (2)

so it decreases quadratically with r

--> This implies that:

1) At r=0, the electric field is 0, because if we substitute r=0 inside eq.(1), we find E(0)=0

2) For r→∞, the electric field tends to zero as well, because according to eq.(2), the electric field strength decreases with the distance r

3) The maximum electric field occur for [tex]r=r_B[/tex], i.e. on the surface of the solid ball: in fact, for [tex]r<r_B[/tex] the electric field increases with distance, while for [tex]r>r_B[/tex] the field decreases with distance, so the maximum value of the field is for [tex]r=r_B[/tex].

The magnitude of the electric field E(r) at a distance r>rb and r from the center of a solid ball with uniform charge density can be calculated using the same formula. The electric field is zero at the center and surface of the ball, and approaches zero as r tends to infinity.

Part A:

The magnitude of the electric field E(r) at a distance r>rb from the center of the ball can be calculated using the formula:

E(r) = (ρ * (4/3) * π * rb³) / (4 * π * ϵ0 * r²)

Part B:

The magnitude of the electric field E(r) at a distance r from the center of the ball can be calculated using the formula:

E(r) = (ρ * (4/3) * π * rb³) / (4 * π * ϵ0 * r²)

Part C:

Based on the formulas provided, the following statements about the electric field are true:

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this is possible because the triceps muscle applies an upward force m perpendicular to the arm, as the drawing indicates. the forearm weighs 20.0 n and has a center of gravity as indicated. the scale registers 118 n. the magnitude of m it run 10 miles.

Force is a parameter which can be used during  pushing or pulling of any object resulting in the object’s interaction or movement, without force the object can not function properly and it can be stopped the direction.

Force is a  quantitative property between two physical bodies, means an object and its environment, there are different types of forces in nature.

If an object in its moving state then that object will be  static or motion, and The external push or pull upon the object called as Force.

The contact force types effort on an object such as Spring Force, Applied Force, Air Resistance Force, Normal Force, Tension Force, Frictional Force

Non-Contact forces are Electromagnetic Force, Gravitational Force, Nuclear Force

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The magnitude of the triceps muscle force M needed to maintain static equilibrium of the forearm is 138 N, which balances the weight of the forearm and the force registered on the scale.

To solve for the magnitude of the force applied by the triceps muscle, M, we must take into account the principles of static equilibrium. In static equilibrium, the sum of all forces acting on the object is zero because it is not moving. Given that the forearm weighs 20.0 N, and the scale registers a downward force of 118 N, we know that there is an upward force required to balance the system and maintain equilibrium.

Let's establish an equation for the vertical forces acting on the forearm:

Since the forearm is in equilibrium, the upward force must balance the downward forces:

M = 138 N

Therefore, the magnitude of the force M applied by the triceps muscle is 138 N.

Answer:

Explanation:

You can always  figure out something to say about a question like this if you have a formula to work with. Likely you do.

There are many ways it can be written

R = k * L / A

So here's the answer.

Resistance = k  which depends of the properties of the material used to make the wire * the  Length of the wire  divided by the cross sectional area of the wire.

The electrical resistance of anything depends on its physical dimensions (length and cross-section area of a wire), and the substance of which it's composed.

In the absence of any external force, the total momentum of a system stays the same is the law of conservation of momentum.

The law of conservation of momentum state that in an isolated system, the total mass of objects does not change which means that the mass of objects before collision is equal to the total mass after collision.

The formula for law of conservation of momentum is. M1V1 + M2v2 = M1V1 + M2V2

M1 is initial mass of objects

V1 initial velocity of the object.

M2 is final mass

V2 final velocity.

Therefore, In the absence of any external force, the total momentum of a system stays the same is the law of conservation momentum.

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Size. The answer is size.

A) The temperature increases by a factor 6

We can use the ideal gas equation:

[tex]pV=nRT[/tex]

where

p is the pressure

V is the volume

n is the number of moles

R is the gas constant

T is the temperature

We can also rewrite it as

[tex]\frac{pV}{T}=nR[/tex]

The gas is in a sealed container - this means the amount of gas is fixed, so n is constant. Since R is constant too, the term on the right in the equation is constant. So we can rewrite the equation as:

[tex]\frac{p_1V_1}{T_1}=\frac{p_2 V_2}{T_2}[/tex]

where in this problem we have:

[tex]V_2 = 2V_1[/tex] (the volume is doubled)

[tex]p_2 = 3 p_1[/tex] (the pressure is tripled)

re-arranging the equation, we find the change in temperature:

[tex]\frac{T_2}{T_1}=\frac{p_2 V_2}{p_1 V_1}=\frac{(3 p_1)(2 V_1)}{p_1 V_1}=6[/tex]

so, the temperature increases by a factor 6.

B) The temperature increases by a factor 1.5.

We can use again the same equation:

[tex]\frac{p_1V_1}{T_1}=\frac{p_2 V_2}{T_2}[/tex]

Where in this case:

[tex]V_2 = \frac{V_1}{2}[/tex] (the volume is halved)

[tex]p_2 = 3 p_1[/tex] (the pressure is tripled)

So, we can find the change in temperature:

[tex]\frac{T_2}{T_1}=\frac{p_2 V_2}{p_1 V_1}=\frac{(3 p_1)(\frac{V_1}{2})}{p_1 V_1}=\frac{3}{2}=1.5[/tex]

So, the temperature increases by 1.5 times.

Considering the combined law equation:

Gay-Lussac's Law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of shocks against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature.

In summary, when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature decreases, gas pressure decreases.

Gay-Lussac's law can be expressed mathematically as follows:

[tex]\frac{P}{T}=k[/tex]

where:

Pressure and volume are related by Boyle's law, which says that the volume occupied by a given mass of gas at constant temperature is inversely proportional to pressure.

Boyle's law is expressed mathematically as:

P× V=k

where:

Finally, Charles's Law consists of the relationship that exists between the volume and the temperature of a certain amount of ideal gas, which is maintained at a constant pressure.

This law says that for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. That is, the volume is directly proportional to the temperature of the gas.

In summary, Charles' law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

[tex]\frac{V}{T}=k[/tex]

where:

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

[tex]\frac{PxV}{T}=k[/tex]

Studying two different states, an initial state 1 and an final state 2, the following will be true:

[tex]\frac{P1xV1}{T1}=\frac{P2xV2}{T2}[/tex]

In this case, you know that:

So, replacing in the combined law equation:

[tex]\frac{P1xV1}{T1}=\frac{3xP1x2xV1}{T2}[/tex]

Solving:

[tex]\frac{P1xV1}{T1}=\frac{6xP1xV1}{T2}[/tex]

[tex]\frac{T2}{T1}=\frac{6xP1xV1}{P1xV1}[/tex]

[tex]\frac{T2}{T1}=6[/tex]

T2= 6×T1

Finally, the temperature increases by a factor of 6.

In this case, you know that:

So, replacing in the combined law equation:

[tex]\frac{P1xV1}{T1}=\frac{3xP1x\frac{V1}{2} }{T2}[/tex]

Solving:

[tex]\frac{P1xV1}{T1}=\frac{3xP1xV1}{2T2}[/tex]

[tex]\frac{T2}{T1}=\frac{3xP1xV1}{2P1xV1}[/tex]

[tex]\frac{T2}{T1}=\frac{3}{2} =1.5[/tex]

T2= 1.5×T1

Finally, the temperature increases by a factor of 1.5.

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Answer:

A moving electric charge creates a magnetic field at all points in the surrounding region.

An electric current in a conductor creates a magnetic field at all points in the surrounding region.

A permanent magnet creates a magnetic field at all points in the surrounding region.

Explanation:

Magnetic field can be produced by:

- moving charges (i.e. a moving electron, or a current in a conductor)

- A magnet

The strength of the magnetic field produced by a current-carrying wire is

[tex]B=\frac{\mu_0 I}{2\pi r}[/tex]

where

I is the current

r is the distance from the wire

As we see from the formula, the magnetic field is produced at all points in the surrounding region, because B becomes zero only when r becomes infinite. The same is true for the magnetic field created by a single moving charge or by a magnet.

The following choices instead are not correct:

- A single stationary electric charge creates a magnetic field at all points in the surrounding region.

- A distribution of electric charges at rest creates a magnetic field at all points in the surrounding region.

Because they involve the presence of stationary charges, and stationary charges do not produce magnetic fields.

The statements that are true concerning the creation of magnetic fields are: A moving electric charge creates a magnetic field; an electric current in a conductor creates a magnetic field; and a permanent magnet creates a magnetic field. Stationary electric charges, regardless if alone or in distribution, do not generate magnetic fields.

Regarding the creation of magnetic fields, the following statements hold true:

Conversely, a single stationary electric charge or a distribution of electric charges at rest do not create magnetic fields - they generate electric fields instead.

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Answer:

[tex]2.74\cdot 10^{14} Hz[/tex]

Explanation:

First of all, let's convert the energy gap from eV to Joules:

[tex]E=1.14 eV \cdot (1.6\cdot 10^{-19}J/eV)=1.82\cdot 10^{-19}J[/tex]

In order to promote the electron to the conduction band, the electron must absorb a photon with an energy at least equal to the energy gap, so:

[tex]E=hf=1.82\cdot 10^{-19}J[/tex]

where

h is the Planck constant

f is the frequency of the photon

Solving for f, we find the lowest frequency needed:

[tex]f=\frac{E}{h}=\frac{1.82\cdot 10^{-19} J}{6.63\cdot 10^{-34}Js}=2.74\cdot 10^{14} Hz[/tex]

The lowest frequency photon that will promote an electron in silicon from the valence band to the conduction band is approximately 1.72 x 10^14 Hz.

To find the lowest frequency photon that will promote an electron in silicon from the valence band to the conduction band, we need to determine the energy difference between the two bands. The energy gap for silicon at 300 K is given as 1.14 eV. Since energy is directly proportional to frequency, we can use the equation E = hf, where E is the energy, h is Planck's constant (6.63 x 10^-34 J s), and f is the frequency. Rearranging the equation gives f = E/h. Plugging in the energy gap for silicon and Planck's constant, we can calculate the lowest frequency photon.

f = (1.14 eV) / (6.63 x 10^-34 J s) = 1.72 x 10^14 Hz

Therefore, the lowest frequency photon that will promote an electron in silicon from the valence band to the conduction band is approximately 1.72 x 10^14 Hz.

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1) Tropical rain forest

2) Temperate deciduous forest

3) Temperate grassland

4) Temperate grassland

In order to find the density of the rock, the student should follow the steps highlighted below.

Steps in determining density of a rock

Weigh the rock using a scale to find out how heavy it is in grams. Ensure that the measurement is very precise.

Find out how much space the rock takes up: There are a few ways to figure out how big an irregularly shaped object, like a rock, is. There are two usual ways:

"Use water to measure volume: Pour water into a cylinder and write down how much there is at the start. " Gently put the rock in the water, and make sure there are no air bubbles stuck.

Find out how much water the rock is pushing out of the way now. Subtract the starting volume from the ending volume to find the volume of the rock.

Archimedes' principle: First, weigh the rock in the air. Then, weigh it again when it's completely underwater. The change in weight when a rock is in water is the same as the weight of the water that the rock pushes aside. Find the volume of the rock by dividing its weight by the density of water, which is usually 1 gram per cubic centimeter.

To find the density of a rock, divide its mass (in grams) by its volume (in cubic centimeters or milliliters). Density = Mass / Volume

Density is the amount of mass in a certain amount of space. It is calculated by dividing the amount of mass by the amount of space it takes up.

Ensure that the measurements for mass and volume are the same (like both in grams or both in kilograms, both in cubic centimeters or both in milliliters).

After calculating the density, write it in the right units. Density is usually measured in grams per cubic centimeter (g/cm³) or kilograms per cubic meter (kg/m³).

Answer:

1.3 m

Explanation:

The work done in lifting the backpack is equal to the change in gravitational potential energy of the backpack, so:

[tex]\Delta U=W \Delta h[/tex]

where

W = 100 N is the weight of the backpack

[tex]\Delta h[/tex] is the change in heigth of the object

In this problem, we know that

[tex]\Delta U = 130 J[/tex]

so we can re-arrange the equation to find the change in height of the backpack:

[tex]\Delta h = \frac{\Delta U}{W}=\frac{130 J}{100 N}=1.3 m[/tex]

If we are talking about simple harmonic motion, we are talking about waves and in this case the frequency [tex]f[/tex] is related to the period of oscillation [tex]T[/tex] in an inverse proportion.

Now, for a mass-on-a-spring system the period is given by:

[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex]

Where [tex]m[/tex] is the oscillating mass and [tex]k[/tex] the spring constant, which depends on the spring stillness.

As we can see in this equation:

If we change [tex]m[/tex] and [tex]k[/tex] we will affect the period, hence the frequency.

Nevertheless, we do not see any relation with the Amplitude, this means the period (hence the frequency) does not depend on the amplitude.

Answer:no Explanation:

It keeps the eggs warm until they hatch is your answer

Answer:

55 mA

Explanation:

Ohm's law states:

V = IR

where V is voltage, I is current, and R is resistance.

220 V = I (4000 Ω)

I = 0.055 A

I = 55 mA

Resistors 'C' and 'D' are in series.  There's only one possible route for current to flow through them.  

Every electron that flows through one of them has to flow through the other one.  

The current (amount of charge per second) must be the same in 'C' and 'D', no matter how many ohms of resistance either one may have. (answer-choice B)

Answer:

C & D

Explanation:

physics s2

Answer:

35.7 mW

Explanation:

The intensity of light after passing through a polarizer is given by

[tex]I=I_0 cos^2 \theta[/tex]

where

[tex]I_0[/tex] is the initial intensity of the light

[tex]\theta[/tex] is the angle between the direction of polarization of the initial light and the transmission axis of the polarizing filter

Keeping in mind that the power is directly proportional to the intensity:

[tex]P \propto I[/tex]

we can rewrite the previous equation as

[tex]P=P_0 cos^2 \theta[/tex]

where we have

[tex]P_0 = 200 mW[/tex]

[tex]\theta=90^{\circ}-25^{\circ}=65^{\circ}[/tex] (because the initial light is horizontally polarized, while the axis of the filter is 25 degrees from the vertical

So, the power of the laser beam emerging from the filter is

[tex]P=(200 mW) cos^2 65^{\circ}=35.7 mW[/tex]

The power of the laser beam as it emerges from the polarizer is approximately 35.7 mW, as calculated using Malus's Law.

The power P' of a light beam after passing through a polarizer can be determined by Malus's Law:

where P is the initial power of the light, θ is the angle between the light's initial polarization direction and the axis of the polarizer.

In this case, the light is horizontally polarized and the axis of the polarizer is 25° from vertical, so the angle θ we need is the complement of 25°, which is 65° (since 90° - 25° = 65°).

Plug in the given values: P = 200 mW and θ = 65° into the equation, we get:

Solving this you get: P' = 35.7 mW

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Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes.

This is where density  [tex]D[/tex]  appears as a  physical characteristic property of matter that establishes a relationship between the mass  [tex]m[/tex] of a body or substance and the volume  [tex]V[/tex] it occupies.

[tex]D=\frac{m}{V}[/tex]

So, according to this equation, the density is inversely ptoportional to the volume:

If the volume increases, the density decreases.

This is what a fish does to have buoyancy, since the density of a body is related to its buoyancy:

A body will float on another fluid if its density is lower.

Answer: The universe is not infinite in space

This question is known as the Olbers' Paradox, who in the 19th century posed a question similar to the following:

If the universe where we live is infinite, static, and uniformly populated with stars (similar to the distribution of trees in an immense forest where we would find a tree in whatever direction we observe), which have eternally existed; the light of all of them would intensely illuminate any point of space.

That is to say, we should see stars in any direction, no matter how far away they are and the celestial vault should be exaggeratedly bright.

But this is not the case, the night sky is dark and the universe too.

Well, although the standard cosmological model of the universe suggests that it is infinite, the observable universe is not.

In other words, the universe is finite.

Then, in a universe of limited size, even having a great quantity of stars and galaxies, all of them would not be enough to illuminate all the space.

In addition, there is another important point: Not only the universe is finite, also its age is; this means it had a beginning.

Hence, having a finite observable universe that is continuously expanding, distant stars and galaxies move away even further.

So, when we look at a star that is 1 million light years away, we are seeing the star as it was seen 1 million years ago.

This means that the amount of light that comes to us from distant stars decreases all the time.

Therefore the light from the most distant stars has not yet had enough time to reach us.

Answer:

It is B.

Explanation:

Answer:

Explanation:

The center of mass lies on a line that joins position 4 of one start with position 4 of the other star.  The shortest distance between these two points will produce the largest velocity. You are using F = m v^2/R

Small R = large force.

Large Force = increased speed.

The masses don't have any effect on the outcome: they remain constant.

Answer:

B) 10 parsecs

Explanation:

The distance of a star measured with the parallax method is given by:

[tex]d=\frac{1}{\theta}[/tex]

where

d is the distance, in Parsec

[tex]\theta[/tex] is the parallax angle, in arcseconds

For the star in the problem, the parallax angle is

[tex]\theta=0.1''[/tex]

therefore, the distance of the star is

[tex]d=\frac{1}{0.1''}=10 pc[/tex]

so, 10 parsecs.

Let:

m = mass (kg)

v = velocity (m/s)

KE = 0.5m(v^2)

KE = 0.5(70)(4^2)

KE = 560 Joules.

The kinetic energy of a 70 kg person running at 4 m/s, calculated using the formula KE = 0.5 * m * v^2, is 560 Joules.

The kinetic energy of a body can be calculated using the formula KE = 0.5 * m * v^2, where KE represents kinetic energy, m represents mass, and v represents velocity. In this case, we have a person with a mass of 70 kg who is running at a speed of 4 m/s. Substituting these values into the formula yields: KE = 0.5 * 70 kg * (4 m/s)^2 = 560 Joules. Hence, the kinetic energy of the person running at 4 m/s is 560 Joules.

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