Please help!!!
Particles q₁ = -8.99 μC, q2 = +5.16 μµC, and
93-89.9 μC are in a line. Particles q₁ and q2 are
separated by 0.220 m and particles q2 and q3 are
separated by 0.330 m. What is the net force on
particle q₁?

The elongation of the rod due to the 1000-kg object hanging from it is approximately 0.0093 meters, or 9.3 millimeters

To calculate the elongation of the steel rod, we'll use the formula:

Elongation = (Force * Length) / (Area * Young's modulus)

First, let's find the force acting on the rod, which is the weight of the object:

Force (F) = mass (m) * acceleration due to gravity (g)
F = 1000 kg * 9.81 m/s²
F = 9810 N

Next, we'll find the area (A) of the rod's cross-section:

Area (A) = π * (diameter/2)²
A = π * (0.008 m / 2)²
A ≈ 5.0265 x 10^-5 m²

Now, we can plug these values into the elongation formula:

Elongation = (9810 N * 5.0 m) / (5.0265 x 10^-5 m² * 210,000 x 10^6 N/m²)
Elongation ≈ 0.0093 m

So the elongation of the rod due to the 1000-kg object hanging from it is approximately 0.0093 meters, or 9.3 millimeters.

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The ratio of the sun's gravitational force on the moon to the earth's gravitational force on the moon is approximately 1:2.

The moon is primarily influenced by the gravitational forces of both the sun and the earth. However, since the sun is much more massive than the earth, its gravitational force on the moon is about 2 times stronger than that of the earth's gravitational force.

Therefore, the ratio of the sun's gravitational force on the moon to the earth's gravitational force on the moon is roughly 1:2.

Understanding the gravitational forces acting on the moon is important in explaining its orbit around the earth and its influence on the tides.      

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The wavelength of the wave propagating down the rope will be 2.43m.

The wavelength of the wave propagating down the rope can be calculated using the formula λ = v/f, where v is the velocity of the wave and f is the frequency.

The velocity of the wave can be found using the formula v = √(T/μ), where T is the tension in the rope and μ is the mass per unit length.

Substituting the given values, we get v = √(2.42/0.628) = 3.05 m/s.

The frequency is given as 2 oscillations per second.

Therefore, the wavelength is λ = 3.05/2 = 1.525m for one complete oscillation. For 2 complete up-and-down oscillations per second, the wavelength will be 2*1.525 = 2.43m.

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If the spectral line of a distant galaxy is broadened, spanning a range of wavelengths, it may indicate that there is a significant amount of movement or turbulence within the gas that is emitting the light.

This could be due to factors such as rotation, outflow, or collisions between gas clouds. The broadening of the spectral line is known as line broadening and can be used to study the dynamics of galaxies and their gas content. It is important to note that other factors, such as instrumental effects, can also contribute to line broadening, so careful analysis and interpretation of the data is necessary to draw accurate conclusions.

What is line broadening ?

The optical spectra of normal stars are continuous spectra overlaid by absorption lines . There are two factors to consider when adding up the spectra of a number of stars to produce the spectrum of a galaxy:

Different types of star have different absorption lines in their spectra. When the spectra are added together, the absorption lines are 'diluted' because a line in the spectrum of one type of star may not appear in the spectra of other types.

Doppler shifts can affect all spectral lines. All lines from a galaxy share the red-shift of the galaxy, but Doppler shifts can also arise from motions of objects within the galaxy. As a result, the absorption lines become broader and shallower.

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The frequency of the incident Ultraviolet light is approximately 7.28 × 10^14 Hz

To find the frequency of the incident ultraviolet light, we can use the relationship between the speed of light, wavelength, and frequency:

c = λ * ν

where c is the speed of light (approximately 3.00 × 10^8 m/s), λ is the wavelength, and ν is the frequency.

Given:

λ = 4.12 × 10^(-7) m

First, we need to convert the work function from electron volts (eV) to joules (J). The conversion factor is 1 eV = 1.602 × 10^(-19) J.

Work function (Φ) = 4.72 eV * (1.602 × 10^(-19) J/eV)

≈ 7.56 × 10^(-19) J

Now, we can rearrange the equation to solve for frequency:

ν = c / λ

ν = (3.00 × 10^8 m/s) / (4.12 × 10^(-7) m)

≈ 7.28 × 10^14 Hz

Therefore, the frequency of the incident ultraviolet light is approximately 7.28 × 10^14 Hz.

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The distance between the crests of a wave with frequency 440Hz that moves through a string under a tension of 41N and with a linear density of 0.75g/m is approximately 1.496 meters.

To find the distance between the crests of the wave, we need to use the wave speed formula:

v = √(T/μ)

where v is the wave speed, T is the tension in the string, and μ is the linear density of the string.

First, we need to convert the linear density from grams per meter to kilograms per meter, since the unit of tension is Newtons:

μ = 0.75 g/m = 0.00075 kg/m

Next, we can plug in the values given in the problem:

v = √(41 N / 0.00075 kg/m) = 658.3 m/s

Now, we can use the frequency of the wave to find its wavelength:

λ = v/f

where λ is the wavelength and f is the frequency.

Again, we can plug in the values given:

λ = 658.3 m/s / 440 Hz = 1.496 m

Therefore, the distance between the crests of the wave is approximately 1.496 meters.

In summary, the distance between the crests of a wave with frequency 440Hz that moves through a string under a tension of 41N and with a linear density of 0.75g/m is approximately 1.496 meters.

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The information provided; it appears that the Andromeda galaxy is moving towards us. This is because the red emission line in the Balmer series of Hydrogen.

The case, the shift is towards the blue end of the spectrum, indicating that the Andromeda galaxy is moving towards us. To calculate the relative speed, we can use the formula v = Δλ/λ * c, where Δλ is the shift in wavelength, λ is the original wavelength, and c is the speed of light. Using the values provided, we get v = 0.66/656 * 3.00 x 10^5 km/s = 302.4 km/s. As a fraction of the speed of light, this is approximately 0.001, or 0.1%. While this may seem like a small percentage, it is important to remember that the Andromeda galaxy is still incredibly far away from us, at a distance of 2.9 million light-years. The fact that we can detect this shift in wavelength at all is a testament to the incredible precision of modern astronomical instruments.

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If 500J of work is required to compress the gas to half its initial volume, then 900J of work is required to compress the gas by a factor of 10.

Relation between work done and change in volume:

The relation between work done, pressure and volume is given by,

Work done = Pressure x Change in Volume

If 500 J of work is required to compress the gas to half its initial volume, we can use the equation:

Work done = Pressure x Change in Volume

Since the temperature is constant, we can assume that the pressure remains constant as well. Therefore, we can write:

500 J = P x (Vi/2 - Vi)

where Vi is the initial volume of the gas and P is the constant pressure.

Simplifying the equation, we get:

500 J = P x (Vi/2)

P = (1000 J/m³) / Vi

Now, we can use the same equation to find the work required to compress the gas by a factor of 10:

Work done = P x Change in Volume

Let's call the final volume Vf. We know that Vf = Vi/10. Therefore:

Work done = P x (Vi - Vf)

Work done = P x (Vi - Vi/10)

Work done = P x (9Vi/10)

Substituting P = (1000 J/m³) / Vi, we get:

Work done = (1000 J/m³) x (9Vi/10) / Vi

Work done = 900 J

Therefore, 900 J of work must be done to compress the gas by a factor of 10, starting from its initial volume.

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To find the total work done in pushing the plate back 1 meter, we need to integrate the force over the displacement:

∫(100(1 cos(10x2))) dx from x=0 to x=1

Simplifying the integrand, we get:

∫(100 cos(10x2)) dx from x=0 to x=1

Integrating, we get:

(10 sin(10) - 10 sin(0)) from x=0 to x=1

= 10(sin(10) - sin(0))

≈ 8.14 Joules

The units of the final answer will be Newton-meters (N·m) since force is in Newtons and displacement is in meters.
Hi! To find the total work done in pushing the plate back 1 meter, we will use the formula for work: W = ∫F dx, where W is the work done, F is the force, and dx is the displacement. In this case, the force F is given by the equation: F = 100(1 cos(10x2)) Newtons.

To set up the integral, we will integrate the force equation with respect to displacement x, over the interval from the starting position (x = 0) to the final position (x = 1 meter).

The integral representing the total work done is:

W = ∫[100(1 cos(20x))] dx, with the limits of integration from x = 0 to x = 1.

The units of the final answer will be Newton-meters (N·m) since force is in Newtons and displacement is in meters.

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The researchers calculated that a typical neutron star had a radius of around 11 km. A neutron star has a density of radius 3 1017 kg/m, which is the same as nuclear matter.

With their latest research, the scientists have discovered proof that the strong nuclear force produces a repulsive force between neutrons at a neutron star's core that prevents the star from collapsing in on itself when particles are arranged in much denser configurations and separated by smaller distances.

The maximum mass of neutron stars has a new upper limit, according to astronomers: It is limited to 2.16 solar masses. We are aware of their little size: A neutron star with a mass 1.4 times that of the sun is predicted to have a radius of between 8 and 16 kilometres.

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According to the given informationthe amplitude of the magnetic field for the light is approximately 2.03 x 10^-11 Tesla.

To find the amplitude of the magnetic field for the light, you'll first need to calculate the amplitude of the electric field. The intensity (I) of light is related to the amplitudes of electric (E) and magnetic (B) fields as follows:

I = (1/2) * c * ε₀ * E²

where c is the speed of light (3 x 10^8 m/s) and ε₀ is the vacuum permittivity (8.85 x 10^-12 F/m).

After finding E, you can find the amplitude of the magnetic field using the relation:

B = E/c

Given the intensity I = 800 W/m², let's find the amplitude of the magnetic field:

1. Solve for E:
800 = (1/2) * (3 x 10^8) * (8.85 x 10^-12) * E²
E ≈ 6.10 x 10^-3 V/m

2. Solve for B:
B = (6.10 x 10^-3) / (3 x 10^8)
B ≈ 2.03 x 10^-11 T

The amplitude of the magnetic field for the light is approximately 2.03 x 10^-11 Tesla.

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In the double-slit experiment with electrons, the electrons arrive at the screen in an interference pattern. The double-slit experiment demonstrates the wave-particle duality of electrons.

When electrons are passed through two slits, they create an interference pattern on the screen behind the slits. This pattern results from the constructive and destructive interference of the electron waves.

Constructive interference occurs when the peaks of two waves align, creating a brighter spot on the screen. Destructive interference happens when the peak of one wave aligns with the trough of another wave, resulting in a darker spot on the screen.
The double-slit experiment highlights the dual nature of electrons as both particles and waves, as they arrive at the screen in an interference pattern.

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Metal mounting yoke not grounded if no equipment grounding conductor and replacement switch wiring does not have one.

According to the National Electrical Code (NEC), the metal mounting yoke of a replacement switch does not need to be connected to an equipment grounding conductor if the wiring at the existing switch does not contain an equipment grounding conductor and the replacement switch wiring does not have one either.

However, it is important to note that if there is an equipment grounding conductor present in the switch box, it must be connected to the metal mounting yoke of the replacement switch.

Failure to properly ground the switch can result in a dangerous electrical shock hazard.

Always consult with a licensed electrician if you are unsure about proper grounding procedures for electrical switches.

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Some advantages of using these locations for solar power include the long hours of sunlight during summer months, low population density, and potential environmental benefits. Disadvantages include the limited sunlight during winter months, harsh weather conditions, high installation costs, and potential challenges in connecting to the power grid.

(Advantages)
Long hours of sunlight during summer: In the summer months, the Arctic Circle experiences 24-hour sunlight, which can result in higher solar energy production during that period.Low population density: The Arctic Circle's low population density means there's plenty of space for large-scale solar installations, minimizing potential land use conflicts.Environmental benefits: Solar power is a clean and renewable energy source, which can contribute to reducing greenhouse gas emissions and help in mitigating climate change.

(Disadvantages)
Limited sunlight during winter months: In the winter months, the Arctic Circle experiences little to no sunlight, making solar power generation extremely limited during that time.Harsh weather conditions: The Arctic Circle's extreme cold, snow, and ice can cause damage to solar panels and other equipment, increasing maintenance costs. High installation costs: Due to the remote location and challenging environment, installation costs for solar power arrays can be significantly higher than in other regions.

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The surface tension of the soap solution is approximately 0.048 N/m.

W = γΔA

where W is the work done, γ is the surface tension of the soap solution, and ΔA is the change in the area of the soap bubble.

We are given W = 2.4 × [tex]10^{-4[/tex] J and ΔA = 100 cm² - 50 cm² = 50 cm² = 5 × [tex]10^{-3[/tex] m².

Substituting the given values in the equation, we get:

2.4 × [tex]10^{-4[/tex] J = γ × 5 × [tex]10^{-3[/tex] m²

Solving for γ, we get:

γ = (2.4 × [tex]10^{-4[/tex] J) / (5 × [tex]10^{-3[/tex] m²)

γ ≈ 0.048 N/m

Surface tension is a physical phenomenon that occurs at the interface between two different phases of matter, such as between a liquid and a gas. It is the measure of the cohesive forces between the molecules at the surface of a liquid, which tend to minimize the surface area and form a surface layer with higher energy than the bulk of the liquid. This results in the formation of a surface film that is capable of supporting objects that are lighter than the liquid itself.

Surface tension arises due to the strong intermolecular forces between the molecules of the liquid. The surface tension of a liquid is influenced by factors such as temperature, pressure, and the presence of other substances in the liquid. Surface tension plays an important role in many natural phenomena, such as the ability of insects to walk on water, the formation of raindrops, and the capillary action in plants.

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Stopping a car moving at twice the original speed requires four times as much work compared to the original speed because the kinetic energy is quadrupled when the speed is doubled.

Stopping a car moving at twice the original speed requires a significant increase in the work done by the brakes. The kinetic energy of the car is proportional to the square of its speed, so if the speed is doubled, the kinetic energy is quadrupled.

To see this more clearly, consider the equation for kinetic energy:

KE = 1/2 * m * v²

where KE is the kinetic energy, m is the mass of the car, and v is its velocity. If the speed of the car is doubled, the kinetic energy becomes:

KE' = 1/2 * m * (2v)² = 2 * (1/2 * m * v²) = 2 * KE

This means that the kinetic energy of the car moving at twice the original speed is twice that of the original speed. To bring the car to a stop, this entire amount of kinetic energy must be dissipated by the brakes, which requires four times as much work compared to the original speed.

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The center of mass of this system is 1.25 AU away from star X and 3.75 AU away from star Y.

We're given that stars X and Y are 5 AU apart, and star X is four times as massive as star Y. We need to find the center of mass for this system.

1. Let's denote the mass of star Y as 'm' and the mass of star X as '4m' (since it is four times as massive).

2. To find the center of mass (COM), we can use the formula: COM = (m1 * d1 + m2 * d2) / (m1 + m2)

3. Let's assume star X is at position 0 AU and star Y is at position 5 AU. So, d1 (distance of star X from COM) is what we're trying to find and d2 (distance of star Y from COM) = 5 AU - d1.

4. Now we can substitute the values in the formula:
COM = (m * (5 - d1) + 4m * d1) / (m + 4m)

5. Simplify the equation:
COM = (5m - md1 + 4md1) / 5m

6. Combine the terms with 'd1':
COM = (5m + 3md1) / 5m

7. Since the center of mass is a fixed point, we can equate it to d1:
d1 = (5m + 3md1) / 5m

8. Solve for d1:
d1 = 5 / (3 + 1) = 5 / 4 = 1.25 AU

9. Now, we can find d2 (distance from star Y to the COM) by subtracting d1 from the total distance of 5 AU:
d2 = 5 AU - 1.25 AU = 3.75 AU

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The velocity of the ball as it passes through its lowest point is about 8.72 m/s.

To find the velocity of the ball as it passes through its lowest point, we can use the principle of conservation of energy. At the highest point, all of the potential energy is converted into kinetic energy when the ball reaches the lowest point.

The potential energy (PE) at the highest point is given by:

PE = m * g * h,

where m is the mass of the ball, g is the acceleration due to gravity, and h is the vertical height relative to the lowest point.

In this case, the height (h) can be calculated as the vertical component of the string length:

h = L * sin(θ),

where L is the length of the string and θ is the angle the string makes with the vertical (75 degrees below the horizontal).

Substituting the given values, we have:

h = 4 m * sin(75 degrees).

Using a calculator, we find:

h ≈ 4 m * 0.96592582628 ≈ 3.86370330512 m.

Now, let's calculate the potential energy at the highest point:

PE = 2 kg * 9.8 m/s^2 * 3.86370330512 m ≈ 76.1209806394 J.

According to the conservation of energy, this potential energy is converted entirely into kinetic energy (KE) at the lowest point.

KE = PE = 76.1209806394 J.

The kinetic energy is given by:

KE = (1/2) * m * v^2,

where v is the velocity of the ball at the lowest point.

Rearranging the equation, we can solve for v:

v^2 = (2 * KE) / m,

v^2 = (2 * 76.1209806394 J) / 2 kg,

v^2 = 76.1209806394 m^2/s^2,

v ≈ √76.1209806394 m^2/s^2,

v ≈ 8.72360824345 m/s.

Therefore, the velocity of the ball as it passes through its lowest point is approximately 8.72 m/s.

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In an electroscope, the leaf is initially deflected when a charged object (such as a rod) is brought close to it. This happens because the charges on the rod induce opposite charges in the leaf, causing it to be attracted to the rod and move away from its original position.

In the experiment involving an electroscope and a rod, the electroscope's leaf goes back to its original position after the rod is removed due to the following reasons:

1. When the charged rod is brought near the electroscope, it induces an opposite charge on the nearest part of the electroscope, causing the electroscope's leaf to repel away from the metal stem.

2. Once the rod is removed, the charges in the electroscope redistribute themselves, returning to a balanced state. This causes the leaf to go back to its original position since there is no longer any net charge to cause repulsion.

In both parts of the experiment, the leaf returns to its original position after the rod is removed because the removal of the rod eliminates the charge imbalance that initially caused the leaf to move.

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The diffraction grating of the light of wavelength 580 nm and a screen is placed 2.5 m away, and the distance between the zero order and first order diffraction spots is 35 cm has 31.5 lines per mm etched on it.

The formula to use in this case is:

d sin θ = mλ

Where d is the distance between adjacent lines on the diffraction grating, θ is the angle between the incident light and the diffracted light, m is the order of the diffraction (0 for the zero order, 1 for the first order, etc.), and λ is the wavelength of the incident light.

In this case, we are given λ = 580 nm, m = 1, d is the unknown we are looking for, and θ can be calculated from the geometry of the setup:

tan θ = opposite / adjacent

= (35 cm / 2) / 2.5 m

= 0.07

θ = arctan(0.07)

= 4 degrees

Now we can rearrange the formula to solve for d:

d = mλ / sin θ

= (1)(580 nm) / sin(4 degrees)

= 0.0317 mm

Finally, we can convert this to lines per mm by taking the reciprocal:

lines per mm = 1 / d

= 31.5 lines per mm

Therefore, the diffraction grating has 31.5 lines per mm etched on it.

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The fundamental frequency of the original tube will be the same as the fundamental frequency of the piece open at both ends, which is 425 Hz.

To find the fundamental frequency of the original tube, we can use the relationship between the fundamental frequency and the length of a tube open at both ends or open at one end.

For a tube open at both ends, the fundamental frequency (f) is given by:

[tex]f = (v / 2L),[/tex]

where:

f is the frequency,

v is the speed of sound in the medium (assuming it's constant),

L is the length of the tube.

For a tube open at one end, the fundamental frequency is given by:

where:

f is the frequency,

v is the speed of sound in the medium,

L is the length of the tube.

Let's assume the lengths of the two shorter pieces of the tube are L1 and L2, with L1 > L2.

Given that the piece open at both ends has a fundamental frequency of 425 Hz (f1 = 425 Hz) and the piece open at one end has a fundamental frequency of 675 Hz (f2 = 675 Hz), we can set up the following equations:

[tex]425 Hz = (v / 2L1),675 Hz = (v / 4L2).[/tex]

We want to find the fundamental frequency of the original tube, so let's express the lengths in terms of the original tube length (L):

[tex]L1 = xL,L2 = (1 - x)L,[/tex]

where x is the proportion of the original length.

Substituting these expressions into the equations, we have:

[tex]425 Hz = (v / 2(xL)),675 Hz = (v / 4((1 - x)L)).[/tex]

Now, we can solve for v in terms of x by rearranging the equations:

v = 850xL Hz,

v = 2700(1 - x)L Hz.

Since the speed of sound (v) is constant, we can equate these expressions:

[tex]850xL Hz = 2700(1 - x)L Hz.[/tex]

Simplifying the equation:

850x = 2700 - 2700x,

3550x = 2700,

x ≈ 0.7606.

Now, we can find the length of the original tube (L):

[tex]L = L1 + L2 = xL + (1 - x)L = 0.7606L + (1 - 0.7606)L = 0.2394L + 0.7606L = L.[/tex]

Therefore, the lengths of the two shorter pieces are proportional to the original length, and the original tube remains unchanged.

As a result, the fundamental frequency of the original tube is 425 Hz.

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The magnitude will be twice the initial velocity, as it has traveled back to its starting point in the opposite direction. The magnitude of the electron's velocity when it returns to its starting point in the opposite direction of its initial velocity can be calculated using the conservation of energy principle. When the electron reaches its maximum displacement from its starting point, it has maximum potential energy and zero kinetic energy.

Assuming the initial velocity of the electron is v, the maximum displacement from the starting point is d, and the electric potential energy of the electron is E, we can write the conservation of energy equation as:
E = (1/2)mv^2 + qVmax = (1/2)mvmax^2
Where m is the mass of the electron, q is its charge, Vmax is the maximum electric potential difference between the starting point and the maximum displacement point, and vmax is the velocity of the electron when it reaches its starting point.
Solving for vmax, we get:
vmax = sqrt(2qVmax/m)
Since the electron returns to its starting point in the opposite direction of its initial velocity, its final velocity will be -v.
|-v| = |-(sqrt(2qVmax/m))| = sqrt(2qVmax/m)
The magnitude of an electron's velocity when it returns to its starting point in the opposite direction of its initial velocity can be calculated using the formula:
v = 2 * u

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The electric field of green light with a wavelength of 546 nm goes through a cycle from maximum to maximum approximately 5.49 x 10¹⁴ times per second.

To determine how many times per second the electric field of green light with a wavelength of 546 nm goes through a cycle from maximum to maximum, we need to find its frequency.

1. We know the wavelength (λ) is 546 nm or 546 x 10⁻⁹ meters.
2. We need to use the speed of light (c) in a vacuum, which is approximately 3 x 10⁸ meters per second.
3. Use the formula relating the speed of light, wavelength, and frequency:

c = λ × f, where f is the frequency.

Now, we can solve for the frequency (f) using the given information:

f = c / λ
f = (3 x 10⁸ m/s) / (546 x 10⁻⁹ m)
f ≈ 5.49 x 10¹⁴ Hz

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Largest to smallest emissivity: 1) black body, 2) dull black surface, 3) shiny metal surface, 4) white surface.

Emissivity is a measure of how much thermal radiation a surface emits compared to a perfect black body.

A black body has the highest emissivity of 1.0, meaning it emits all the radiation it can at a given temperature.

Next is a dull black surface, with an emissivity of around 0.9.

A shiny metal surface has an emissivity of around 0.1, because it reflects a lot of the radiation it receives.

Finally, a white surface has the smallest emissivity of around 0.05, because it reflects most of the radiation and emits very little.

Therefore, the ranking from largest to smallest emissivity of the surfaces in question is: 1) black body, 2) dull black surface, 3) shiny metal surface, 4) white surface.

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The correct answer is D) not change if you double pressure on surface of can on water where buoyant force will be applied.

The buoyant force on a stone submerged in water depends on the volume of the displaced water and the density of the water, according to Archimedes' principle. The equation for buoyant force (F_b) is:

[tex]F_b = ρ * V * g[/tex]
where ρ is the density of the fluid, V is the volume of the displaced fluid, and g is the acceleration due to gravity.

Doubling the pressure on the surface of the can of water does not change the volume of water displaced by the stone or the density of the water. Therefore, the buoyant force on the stone remains the same, even if the pressure on the surface of the water is doubled.

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The Marine's power output while climbing the 10.0m vertical rope at a constant speed in 8.00s is: 875 watts.

To calculate the power output of the 700-N Marine climbing a 10.0m vertical rope at a constant speed in 8.00s, we will use the following terms and formula:

1. Weight (force): The Marine's weight is given as 700 N.

2. Distance: The Marine climbs a vertical rope of 10.0 m in height.

3. Time: The Marine takes 8.00 s to complete the climb.

4. Power: This is the output we want to determine.

The formula for power is:

Power = (Weight × Distance) / Time

Using the given information, we can now calculate the power output:

Power = (700 N × 10.0 m) / 8.00 s

Power = 7000 Nm / 8.00 s

Power = 875 W (watts)

Thus, the Marine's power output while climbing the 10.0m vertical rope at a constant speed in 8.00s is 875 watts.

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Complete question:

A 700- N Marine in basic training climbs a 10.0m vertical rope at a constant speed in 8.00s. What is his power output?

Momentum is conserved in an isolated system, i.e., a system that does not interact with its surroundings.

In such a system, the total momentum of all the objects in the system remains constant. This is known as the law of conservation of momentum.

In a nondissipative system, i.e., a system where there is no energy loss due to friction or other dissipative forces, momentum is also conserved. This is because momentum is a conserved quantity, and in the absence of any external forces, the total momentum of a system remains constant.

In a system with only conservative forces, momentum is also conserved. Conservative forces are forces that depend only on the position of an object and not on its velocity or acceleration. Examples of conservative forces include gravitational and electric forces. These forces do not dissipate energy and, therefore, do not affect the conservation of momentum.

However, in systems with non-conservative forces, such as friction or air resistance, momentum is not conserved. In such systems, the total momentum of the objects in the system changes as a result of the external forces acting on them.

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When a star emitting visible wavelengths of light is moving toward your telescope, the light will be blueshifted due to the Doppler effect.

The Doppler effect is a phenomenon where the frequency of a wave changes due to the relative motion between the source and the observer.

In the case of a star moving toward your telescope, its light waves are compressed, causing the observed wavelengths to become shorter.

This results in a shift toward the blue end of the visible light spectrum, which is called blueshift. Conversely, if the star were moving away from your telescope, the light would be redshifted, with wavelengths becoming longer and shifting toward the red end of the spectrum.

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When a 1000 kg pallet of bricks is being lowered to the ground by a crane at a constant speed, we can calculate the work done by the cable on the bricks as they descend 11 meters.

In this scenario, the force exerted by the cable is equal to the gravitational force acting on the bricks, as they are moving at a constant speed. The gravitational force can be calculated using the formula F = m * g, where F is the force, m is the mass (1000 kg), and g is the acceleration due to gravity (approximately 9.81 m/s²).

F = 1000 kg * 9.81 m/s² = 9810 N

Now that we have the force, we can calculate the work done by the cable using the formula W = F * d * cosθ, where W is the work, F is the force (9810 N), d is the distance (11 m), and θ is the angle between the force and the direction of movement. In this case, the force and movement direction are both vertically downwards, making θ = 0 degrees, and cosθ = 1.

W = 9810 N * 11 m * 1 = 107910 J

Therefore, the work done by the cable on the bricks as they descend 11 meters is 107,910 Joules.

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Astronomers believe that the reason for the higher abundance of lithium in cosmic rays compared to the Sun and stars is due to cosmic ray spallation.

Cosmic rays, which are high-energy particles that originate from outside the Solar System, can interact with interstellar matter and break apart heavier elements into lighter ones, including lithium. Since the Sun and stars have much stronger magnetic fields and denser atmospheres than the vast regions of interstellar space where cosmic rays travel, they are not as susceptible to cosmic ray spallation. Therefore, the relatively low abundance of lithium in the Sun and stars compared to cosmic rays is thought to be due to this process.

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