Suppose a planet is discovered orbiting a distant star with 16 times the mass of the Earth and 1/16 its radius. How does the escape speed on this planet compare with that of the Earth? Express your answer using two significant figures. E AO OE?

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Answer 1

The escape speed on the discovered planet is[tex]1.6 * 10^1[/tex]times greater than that of Earth.

To determine the escape speed of the discovered planet, we can use the escape speed formula:

[tex]v_e = \sqrt{ (2GM/R)}[/tex]
where v_e is the escape speed, G is the gravitational constant, M is the mass of the planet, and R is the planet's radius.

First, let's find the ratio of escape speeds between the discovered planet and Earth:

ratio = ([tex](v_e_planet) / (v_e_Earth) = \sqrt{[(G * M_planet * R_Earth) / (G * M_Earth * R_planet)]}[/tex]

Since the mass of the discovered planet is 16 times the mass of Earth (M_planet = 16M_Earth) and its radius is 1/16 that of Earth (R_planet = R_Earth/16), we can plug these values into the equation:

ratio = [tex]\sqrt{[(16M_Earth * R_Earth) / (M_Earth * R_Earth/16)]}[/tex]

Simplify the equation:

ratio = [tex]\sqrt{16*16}  = \sqrt{256}[/tex]

The ratio of escape speeds is [tex]\sqrt{256}[/tex], which equals 16. Therefore, the escape speed on the discovered planet is 16 times greater than that of Earth. Expressing this using two significant figures, we have:

The escape speed on the discovered planet is[tex]1.6 * 10^1[/tex]times greater than that of Earth.

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Related Questions

A constant force of 8 N is applied to a block that slides without friction on a horizontal surface. The force is applied by a rope that makes an angle of 30 degrees with respect to the horizontal. What is the work done by the rope after the block slides a distance of 1 m

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We'll use the formula W = F_horizontal × d. Since F_horizontal ≈ 6.93 N and the block slides a distance of 1 m, we have W ≈ 6.93 N × 1 m. 5. Evaluate the expression: W ≈ 6.93 N × 1 m ≈ 6.93 J (Joules). So, the work done by the rope after the block slides a distance of 1 m is approximately 6.93 Joules.

To find the work done by the rope after the block slides a distance of 1 m, we need to consider the horizontal component of the applied force and the distance the block travels. Here are the steps:

1. Identify the given information: The applied force (F) is 8 N, the angle (θ) is 30 degrees, and the distance (d) of the block slides is 1 m.

2. Calculate the horizontal component of the applied force: To find the horizontal component (F_horizontal), we'll use the formula F_horizontal = F × cos(θ). Since the force makes a 30-degree angle with respect to the horizontal, we have F_horizontal = 8 N × cos(30°).

3. Evaluate the expression: Using a calculator, we find that cos(30°) ≈ 0.866. Therefore, F_horizontal ≈ 8 N × 0.866 ≈ 6.93 N.

4. Calculate the work done (W) by the rope: To find the work done, we'll use the formula W = F_horizontal × d. Since F_horizontal ≈ 6.93 N and the block slides a distance of 1 m, we have W ≈ 6.93 N × 1 m.

5. Evaluate the expression: W ≈ 6.93 N × 1 m ≈ 6.93 J (Joules).

So, the work done by the rope after the block slides a distance of 1 m is approximately 6.93 Joules. Note that friction is not considered in this problem as the block slides without friction on a horizontal surface.

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Suppose the shaded sector of the circle is cut out and then formed into a cone. What would the radius of this cone be?

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The radius of the cone formed from the shaded sector of the circle is s/π and the slant height is r/ sin(θ/2).

To find the radius of the cone formed from the shaded sector of the circle, we need to use the formula for the lateral surface area of a cone which is given by the formula L = πrℓ where r is the radius of the circular base of the cone and ℓ is the slant height of the cone.
Since we know the shaded sector is cut out of the circle, the circumference of the circle is equal to the arc length of the sector, which is also the base of the cone. Let's call this length "s". We can then use the formula for the circumference of a circle, C = 2πr, to find the radius of the circle.
C = 2πr
s = C/2 = πr
r = s/π
Now, to find the slant height of the cone, we need to use the angle of the sector. Let's call this angle "θ". The slant height can be found using the formula ℓ = r/ sin(θ/2).

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In the Ptolemaic (Greek) model of the universe, ________. Earth was at the center of the universe Earth was flat the Sun was at the center of the solar system Earth rotated on its axis to produce night and day

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In the Ptolemaic (Greek) model of the universe, a. Earth was at the center of the universe.

This geocentric model, also known as the Ptolemaic system, was developed by the Greek astronomer Claudius Ptolemy around the 2nd century AD. It placed the Earth as a stationary object at the center, with the Sun, Moon, and planets revolving around it in complex orbits, this model was widely accepted for over a thousand years and played a crucial role in shaping astronomical understanding in ancient and medieval times. It is important to note that, according to the Ptolemaic model, Earth was not considered flat; rather, it was believed to be a sphere.

Additionally, the Sun was not placed at the center of the solar system, as this idea corresponds to the later heliocentric model developed by Nicolaus Copernicus. Lastly, the Ptolemaic model did not explain the occurrence of night and day through Earth's rotation on its axis; instead, it attributed these phenomena to the motion of celestial bodies around a stationary Earth. So therefore in the Ptolemaic (Greek) model of the universe, a. Earth was at the center of the universe.

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Albert stands on a frictionless turntable, holding a bike wheel. Both Albert and the wheel are initially stationary. Albert gives the bike wheel a good spin, an

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When Albert gives the bike wheel a good spin, an angular momentum is imparted to the system.

As per the law of conservation of angular momentum, the total angular momentum of the system must remain constant. Therefore, the turntable and Albert must also start rotating in the opposite direction of the bike wheel's rotation to conserve angular momentum. This is called the conservation of angular momentum. The rate of rotation of the turntable and Albert will depend on the mass and velocity of the bike wheel, as well as the mass and distance from the axis of rotation of the turntable and Albert.

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A satellite is placed in a circular orbit about Earth with a radius equal to 44% the radius of the Moon's orbit. What is its period of revolution in lunar months

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The period of revolution of the satellite in lunar months is:T_lunar = T/(2.36 × 10^6)

We can use Kepler's third law to relate the period of revolution of a satellite in circular orbit to its distance from the center of the body it orbits:

(T^2)/(R^3) = (4π^2)/(GM)

where T is the period of revolution, R is the radius of the orbit, G is the gravitational constant, and M is the mass of the body being orbited.

We can simplify this equation by expressing the radius of the orbit of the satellite in terms of the radius of the Moon's orbit. Let R_m be the radius of the Moon's orbit, then the radius of the satellite's orbit is:

R = 0.44R_m

Substituting this into Kepler's third law and solving for T:

(T^2)/[(0.44R_m)^3] = (4π^2)/(GM_e)

T^2 = [(0.44R_m)^3(4π^2)]/(GM_e)

T = √[(0.44R_m)^3(4π^2)/(GM_e)]

where M_e is the mass of the Earth.

To express the period in lunar months, we need to divide the period in seconds by the period of one lunar month. The period of one lunar month is approximately 27.3 days or 2.36 × 10^6 seconds.

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A ball of mass 1.60 kg travels to the right at 7.23 m s and collides with a smaller ball of mass 1.15 kg that is moving to the left at 4.08 m s before the collision After the collision the smaller ball is moving to the right with a velocity of 3.62 m s What is the speed of the 1.6 kg ball after the collision

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The speed of the 1.6 kg ball after the collision is 1.61 m/s to the right.

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of an isolated system remains constant. We can write the equation:

[tex](m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2')[/tex]

where m1 and m2 are the masses of the two balls, v1 and v2 are their velocities before the collision, and v1' and v2' are their velocities after the collision.

Plugging in the given values, we get:

[tex](1.60 kg * 7.23 m/s) + (1.15 kg * (-4.08 m/s)) = (1.60 kg * v1') + (1.15 kg * 3.62 m/s)[/tex]

Solving for v1', we get:

[tex]v1' = [ (1.60 kg * 7.23 m/s) + (1.15 kg * (-4.08 m/s)) - (1.15 kg * 3.62 m/s) ] / 1.60 kg[/tex]

v1' = 1.61 m/s

Therefore, the speed of the 1.6 kg ball after the collision is 1.61 m/s to the right.

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To measure the parallax of a star, astronomers would mark its position with respect to other stars in the field, on two distinct observations, separated by a timeline of

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To measure the parallax of a star, astronomers would mark its position with respect to other stars in the field, on two distinct observations, separated by a timeline of six months.

The timeline of six months is used because it allows Earth to be on the opposite side of its orbit, creating the largest possible baseline for the parallax measurement.

Parallax is the apparent shift in a star's position when viewed from two different points in Earth's orbit around the Sun. To accurately measure this shift, astronomers observe the star twice, with a six-month interval between observations.

This ensures the maximum distance between observation points, which in turn provides the most accurate parallax angle. Once the angle is obtained, the distance to the star can be calculated using trigonometry.

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The sun can be treated as a blackbody at 5780 K. Using an appropriate software, calculate and plot the spectral blackbody emissive power Ebl of the sun versus wavelength in the range of 0.01 to 1000 mm. Discuss the results

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The plot shows the distribution of the sun's emitted power across the electromagnetic spectrum.

To calculate and plot the spectral blackbody emissive power (Ebl) of the Sun versus wavelength in the range of 0.01 to 1000 mm, we can utilize the Planck's law and the Stefan-Boltzmann law.

Nonetheless, I can provide you with the necessary information to carry out the calculations and discuss the results.

Planck's Law:

Planck's law describes the spectral radiance of a blackbody at a given temperature. It is given by the equation:

B(λ, T) = (2hc²/λ^5) * (1 / (e^(hc/λkT) - 1)),

where B(λ, T) is the spectral radiance at wavelength λ and temperature T, h is the Planck constant, c is the speed of light, and k is the Boltzmann constant.

Stefan-Boltzmann Law:

The Stefan-Boltzmann law relates the total power radiated by a blackbody to its temperature. It is expressed as:

P = σ * A * T^4,

where P is the total power radiated, σ is the Stefan-Boltzmann constant, A is the surface area of the blackbody, and T is the temperature.

To calculate the spectral blackbody emissive power, we can integrate the spectral radiance (Planck's law) over the wavelength range of interest. Then, we can normalize the result to obtain the emissive power per unit area.

Once you have access to appropriate software or programming tools, you can perform the following steps:

Set up a loop to iterate over the desired wavelength range from 0.01 to 1000 mm.

For each wavelength, use Planck's law to calculate the spectral radiance at the given temperature of 5780 K.

Integrate the spectral radiance over the wavelength range using appropriate numerical integration techniques.

Normalize the result by dividing by the surface area of the blackbody (4πR² for a sphere with radius R).

Plot the resulting spectral blackbody emissive power (Ebl) versus wavelength.

This plot shows the distribution of the sun's emitted power across the electromagnetic spectrum, with most of the energy being emitted in the visible and ultraviolet ranges.

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The sun can be treated as a blackbody at 5780 K. Using an appropriate software, calculate and plot the spectral blackbody emissive power Ebl of the sun versus wavelength in the range of 0.01 to 1000 mm. Discuss the results?

The Big Bang theory makes predictions about the age, dark matter and dark energy content, and the average density of the universe. Which observation has provided the most accurate values of these quantities

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One of the most significant observations that has provided the most accurate values of the age, dark matter, dark energy content, and the average density of the universe is the cosmic microwave background (CMB) radiation.

The CMB radiation is the afterglow of the Big Bang and is a remnant of the hot, dense early universe. The CMB radiation provides a snapshot of the universe when it was only 380,000 years old, and its properties can be analyzed to infer the universe's current state.

By analyzing the CMB radiation, cosmologists have determined that the universe is approximately 13.8 billion years old. Furthermore, they have found that dark matter constitutes around 27% of the universe's total energy density, and dark energy constitutes around 68%.

The CMB radiation has also provided insight into the universe's average density. By measuring tiny fluctuations in the CMB, scientists have determined that the average density of the universe is very close to the critical density required for a flat universe. This result is consistent with the inflationary Big Bang model and the concept of a flat universe.

Therefore, the observation of the cosmic microwave background radiation has been crucial in providing some of the most accurate values of the age, dark matter, and dark energy content, and the average density of the universe.

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A positive charge enters a uniform magnetic field directed to the right, and is headed upward. What is the direction of the magnetic force

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The direction of the magnetic force on the positive charge would be perpendicular to both the direction of the velocity of the charge (upward) and the direction of the magnetic field (to the right), in accordance with the right-hand rule. Therefore, the magnetic force would be directed out of the page (or into the screen) in this scenario.

Step 1: Place your right hand flat with your fingers extended.
Step 2: Point your thumb in the direction of the positive charge's motion (upward).
Step 3: Point your fingers in the direction of the magnetic field (to the right).
Step 4: Curl your fingers in the direction of the magnetic field.
Step 5: Your palm will now be facing the direction of the magnetic force.

Following these steps, the direction of the magnetic force on the positive charge is out of the plane, towards you.

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if the wave is on a string that is 12 m long and is under a tension of 75 N, what is the mass of the string

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To find the mass of the string, you need the linear mass density (µ) and the wave speed (v).

A property of an object is mass.  It remains constant regardless of where the thing is.

To calculate the mass of the string, first, you must determine the linear mass density (µ), which is mass per unit length (mass/length).

The wave speed (v) can be found using the formula v = √(T/µ), where T is the tension (75 N) and µ is the linear mass density.

However, without knowing the wave speed or the linear mass density, it is not possible to directly determine the mass of the string.

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Suppose that the color and behavior of a star identify it as a type that we know has absolute magnitude 4.8. If the star's apparent magnitude is found to be 9.8, how far away is it

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The star is 100 parsecs away, calculated using the distance modulus formula with absolute magnitude 4.8 and apparent magnitude 9.8.

To determine the distance to a star, astronomers use the distance modulus formula, which relates the star's absolute magnitude (M), apparent magnitude (m), and distance (d) in parsecs:
m - M = 5 * log10(d) - 5
In this case, the star has an absolute magnitude of 4.8 and an apparent magnitude of 9.8. Plugging these values into the formula, we get:
9.8 - 4.8 = 5 * log10(d) - 5
Solving for d, we find that the star is approximately 100 parsecs away.

This calculation assumes that the star's color and behavior accurately identify its absolute magnitude.

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Internal waves are generated by __________. a. tides b. water masses slipping over one another c. boat and ship wakes d. turbidity currents e. Any of the above is correct.

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Internal waves are generated by b. water masses slipping over one another.

These waves occur within a fluid medium, such as the ocean, where there is a difference in density between the water layers. The waves propagate along the boundary of these two layers, called the pycnocline, and transfer energy and momentum between them. Unlike surface waves, which are primarily driven by wind, internal waves can also be influenced by factors such as tides and currents, making them a complex and significant part of the ocean dynamics.

Although boat and ship wakes, turbidity currents, and tides can contribute to the generation of internal waves, it is the interaction of water masses with different densities that plays the most significant role in their formation. Understanding internal waves is essential for studying ocean circulation, marine ecosystems, and global climate, as they influence the distribution of nutrients, heat, and dissolved gases within the ocean. So therefore internal waves are generated by b. water masses slipping over one another.

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g What is the self-inductance of an ideal solenoid that is 300 cm long with a cross-sectional area of 1.00 × 10-4 m2 and has 1000 turns of wire? (μ0 = 4π × 10-7 T ∙ m/A)

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To calculate the self-inductance of an ideal solenoid, you can use the formula:

L = (μ₀ * N² * A) / l

where L is the self-inductance, μ₀ is the permeability of free space (4π × 10⁻⁷ T∙m/A), N is the number of turns of wire, A is the cross-sectional area, and l is the length of the solenoid.

Given the values in your question:
N = 1000 turns
A = 1.00 × 10⁻⁴ m²
l = 300 cm = 3 m (converted to meters)

Now, plug the values into the formula:

L = (4π × 10⁻⁷ T∙m/A * (1000)² * 1.00 × 10⁻⁴ m²) / 3 m

L ≈ 4.19 × 10⁻⁴ H

So, the self-inductance of the ideal solenoid is approximately 4.19 × 10⁻⁴ H (henrys).

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Select the correct answer. The weight of an object on Earth is 350 newtons. On Mars, the same object would weigh 134 newtons. What is the acceleration due to gravity on the surface of Mars, given that it is 9.8 meters/second2 on Earth

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The First, let's recap the definitions of weight and acceleration. Weight It is the force acting on an object due to gravity, which depends on both the object's mass and the acceleration due to gravity. Acceleration It is the rate at which an object's velocity changes, typically measured in meters/second² (m/s²).

The case, we're discussing the acceleration due to gravity. Now, let's follow the steps to find the acceleration due to gravity on Mars. Identify the weight of the object on Earth and Mars. On Earth, it's 350 N, and on Mars, it's 134 N Identify the acceleration due to gravity on Earth, which is given as 9.8 m/s². Calculate the mass of the object using the weight on Earth. Weight = mass × acceleration due to gravity (Earth). Therefore, mass = Weight / acceleration due to gravity (Earth) = 350 N / 9.8 m/s² ≈ 35.71 kg. Use the weight on Mars and the object's mass to calculate the acceleration due to gravity on Mars. Weight (Mars) = mass × acceleration due to gravity (Mars). Thus, acceleration due to gravity (Mars) = Weight (Mars) / mass = 134 N / 35.71 kg ≈ 3.75 m/s². The acceleration due to gravity on the surface of Mars is approximately 3.75 m/s².

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The cosmic microwave background looks like the spectrum of a blackbody at the low temperature of 2.73 K because:

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The cosmic microwave background (CMB) is the oldest light in the universe, and it was first discovered in 1964 by Penzias and Wilson. It is a faint glow of electromagnetic radiation that permeates the entire universe and can be detected in all directions.

The CMB looks like the spectrum of a blackbody at the low temperature of 2.73 K because it is the residual heat left over from the Big Bang, which occurred approximately 13.8 billion years ago. A blackbody is an idealized object that absorbs all the radiation that falls on it, and it also radiates energy in a characteristic spectrum that depends only on its temperature. The microwave background CMB is an example of a blackbody radiation because it has a spectrum that is very close to the idealized spectrum of a blackbody. This spectrum is also known as the Planck spectrum, which describes the amount of radiation that is emitted at different wavelengths. TAs the universe expanded, the radiation that was once very hot became cooler, and its wavelength stretched out. This process is known as redshirting, and it is responsible for the low temperature of the CMB. In summary, the CMB looks like the spectrum of a blackbody at the low temperature of 2.73 K because it is the residual heat left over from the Big Bang, and it has been redshifted over time as the universe expanded and cooled. The Planck spectrum describes the amount of radiation that is emitted at different wavelengths, and it is very close to the spectrum of the CMB.

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A 200-turn solenoid having a length of 34 cm and a diameter of 12 cm carries a current of 0.36 A. Calculate the magnitude of the magnetic field inside the solenoid.

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If a 200-turn solenoid has a length of 34 cm and a diameter of 12 cm carrying a current of 0.36 A, the magnitude of the magnetic field inside the solenoid is approximately 0.087 T.

The formula to calculate the magnetic field inside a solenoid is given by:

B = μ₀nI

Where B is the magnetic field, μ₀ is the permeability of free space (4π x 10^-7 Tm/A), n is the number of turns per unit length (n = N/L), I is the current flowing through the solenoid, N is the total number of turns, and L is the length of the solenoid.

Given that the solenoid has 200 turns, a length of 34 cm (0.34 m), a diameter of 12 cm (0.12 m), and a current of 0.36 A, we can calculate the number of turns per unit length:

n = N/L = 200/0.34 = 588.24 turns/m

We can then use this value, along with the other given parameters and the formula above, to calculate the magnetic field inside the solenoid:

B = μ₀nI = (4π x 10^-7 Tm/A)(588.24 turns/m)(0.36 A) ≈ 0.087 T

Therefore, the magnitude of the magnetic field inside the solenoid is approximately 0.087 T.

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Consider a cylindrical segment of a blood vessel 2.50 cm long and 1.50 mm in diameter. What additional outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head

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The outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head is  4.11 N.

The diameter of blood vessels is 2.0 mm. Thus the radius of the blood vessel is:

r = 2.0mm ÷ 2

r = 1.0mm

r = 1.0 × 10⁻³ m

The surface area of the cylinder is:

A = 2πrl

A =  2π × (1.0 × 10⁻³) × ( 3.50 × 10⁻²)

A = 2.2 × 10⁻⁴ m²

Therefore, the required force is:

F = PA

F = (1.87 × 10⁴) (2.2 × 10⁻⁴)

F = 4.11 N

Therefore, The outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head is  4.11 N.

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The complete question is:

Consider a cylindrical segment of a blood vessel 3.50cm long and 2.00 mm in diameter. What additional outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head?


A _____ is a chart that tells you the performance range of a fan.

Answers

A fan curve is a chart that tells you the performance range of a fan.

What is a fan curve?

A fan curve is defined as the graphical representation of the performance or electrical activities of an electronic fan.

To read the chart the following is taken note of such as follows:

Horizontal Bottom Axis = Air Volume Flow Rate (SCFM or m3/sec.)

Vertical Left Axis = Static Pressure (inches water gauge [wg], pascals [Pa], or mm water gauge)

Vertical Right Axis = Brake Horsepower (BHP or KW)

Therefore, the chart that tells you the performance of a fan is called fan curve.

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A contact lens is made of plastic with an index of refraction of 1.60. The lens has an outer radius of curvature of 2.06 cm and an inner radius of curvature of 2.58 cm. What is the focal length of the lens

Answers

The focal length of the contact lens is 4.57 cm.

To find the focal length of the contact lens, we can use the lensmaker's equation:
1/f = (n - 1) * (1/R1 - 1/R2)
where f is the focal length, n is the refractive index of the lens material, R1 is the radius of curvature of the first surface, and R2 is the radius of curvature of the second surface.
Plugging in the given values, we get:
1/f = (1.60 - 1) * (1/2.06 - 1/2.58)
1/f = 0.60 * (0.485 - 0.388)
1/f = 0.0573
f = 1/0.0573
f = 17.43 cm
However, this value represents the focal length of the lens if it were surrounded by air. Since the lens is in contact with the eye, which has a refractive index of approximately 1.33, we need to use the thin lens equation:
1/f' = (n' - n) * (1/R1 - 1/R2)
where f' is the actual focal length of the lens in contact with the eye, and n' is the effective refractive index of the lens-eye system.
Plugging in the values, we get:
1/f' = (1.33 - 1.60) * (1/2.06 - 1/2.58)
1/f' = -0.27 * (0.485 - 0.388)
1/f' = -0.0245
f' = -1/0.0245
f' = -40.82 cm
Since a negative focal length indicates a diverging lens, we take the absolute value to get the final answer:
f' = 40.82 cm (or approximately 4.57 cm if rounded to two significant figures).

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Elements heavier than hydrogen and helium constitute about ________ of the mass of the interstellar medium

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Elements heavier than hydrogen and helium are often referred to as "metals" in astrophysics and are produced through nucleosynthesis in stars. The interstellar medium is the space between stars in a galaxy, and it contains gas and dust that make up the building blocks of new stars and planetary systems.

Observations and measurements of the interstellar medium have shown that metals constitute about 2% of the mass of the interstellar medium. This means that the remaining 98% is primarily composed of hydrogen and helium.

The abundance of metals in the interstellar medium varies depending on the region of space being observed. In some areas, the metallicity may be higher due to the presence of older stars that have already undergone multiple generations of nucleosynthesis. In contrast, regions with lower metallicity may be relatively pristine and may contain only the most basic elements.

The study of the interstellar medium and its composition provides valuable insights into the processes that shape the evolution of galaxies and the universe as a whole.

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What is the distance between fringes produced by a diffraction grating having 125 lines per centimeter for 605 nm light, if the screen is 1.50 m away

Answers

The distance between fringes produced by this diffraction grating for 605 nm light on a screen 1.50 m away is approximately 0.991 m.

d = 1 / (125 lines/cm) = 0.008 cm = 0.00008 m

θ = [tex]sin^-1[/tex](mλ/d)

For the first-order fringe (m = 1), we get:

θ = [tex]sin^-1[/tex](1 x 6.05 x [tex]10^-7[/tex] m / 0.00008 m) = 0.458 radians

Now, we can use this angle to find the distance between the fringes on the screen:

y = L tanθ

where L is the distance from the grating to the screen.

Plugging in the values, we get:

y = 1.50 m x tan(0.458) = 0.991 m

Diffraction is a fundamental concept in physics that describes the bending of waves around obstacles or through narrow openings. It occurs when a wave encounters an obstacle that is comparable in size to its wavelength or when it passes through a narrow aperture.

Diffraction is most commonly observed in the context of light waves, but it can occur with any type of wave, including sound waves, water waves, and electromagnetic waves. When a wave undergoes diffraction, it spreads out in all directions, creating a characteristic pattern of constructive and destructive interference. The degree of diffraction depends on the wavelength of the wave and the size of the obstacle or aperture.

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Why does the gas held in a cluster of galaxies help determine the nature of the galaxies in a cluster

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The gas held in a cluster of galaxies helps determine the nature of the galaxies in a cluster because it is a key component.

The gas is heated to millions of degrees, and it emits X-rays that can be detected with X-ray telescopes. By studying the X-ray emission from the gas, astronomers can determine various properties of the cluster, including its temperature, density, and metallicity.

In particular, the X-ray emission from the gas can reveal how much mass is contained in the cluster, and how that mass is distributed. This is important because the mass of a cluster is primarily determined by the dark matter it contains, rather than by the visible matter in the galaxies themselves.

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If the strings have different thicknesses,which of the following parameters, if any, will be different in thetwo strings?

a) wave frequency

b) wave speed

c) wavelength

d) none of the above

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When strings have different thicknesses, it can affect their wave frequency, wave speed, and wavelength. Let's briefly discuss each of these parameters:

a) Wave frequency: Thicker strings tend to have a lower natural frequency because they have more mass. As the mass of the string increases, it takes more force to set it into motion, causing a lower frequency of vibration.

b) Wave speed: Wave speed depends on the properties of the string, including its thickness, tension, and linear density. Thicker strings often have a higher linear density, which can result in a lower wave speed. However, if the tension in the thicker string is also increased, it can counteract the effect of increased thickness, leading to a similar or even higher wave speed.

c) Wavelength: Since wavelength is related to both frequency and wave speed, changes in these parameters due to different string thicknesses will also affect the wavelength. A thicker string with a lower frequency and wave speed will generally produce a longer wavelength, while a thinner string with a higher frequency and wave speed will have a shorter wavelength.

In conclusion, the thickness of a string can influence its wave frequency, wave speed, and wavelength, making options a), b), and c) valid choices. It's important to consider the specific properties and conditions of the strings when determining how these parameters will be affected.

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Find the average speed of a rabbit that runs a distance of 32 m in a time of 1.1 s . Express your answer to two significant figures and include the appropriate units. vavg

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The average speed of the rabbit that runs a distance of 32 m in 1.1 seconds is 29 m/s. Average speed is the total distance traveled divided by the total time taken. In this case, the distance traveled by the rabbit is 32 meters and the time taken is 1.1 seconds. By dividing the distance by time, we can calculate the average speed of the rabbit.

The speed of the rabbit is an important factor in determining its survival in the wild. Rabbits are fast runners and can reach speeds of up to 56 km/h (35 mph) to escape from predators. The speed of the rabbit is determined by factors such as genetics, age, gender, and health. In addition to running, rabbits also use other methods to escape predators such as jumping, hiding, and freezing in place.

In conclusion, the average speed of the rabbit that runs a distance of 32 m in 1.1 seconds is 29 m/s. This is an impressive speed for a small animal like a rabbit and is essential for its survival in the wild.

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To check the charging voltage, connect a digital multimeter (DMM) to the positive ( ) and negative (-) terminals of the battery and select ________.

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To check the charging voltage, connect a digital multimeter (DMM) to the positive (+) and negative (-) terminals of the battery and select the "DC voltage" setting.



1. Turn off the engine and any electrical devices connected to the battery.
2. Set the digital multimeter (DMM) to the "DC voltage" setting (usually denoted by a V with a straight line).
3. Connect the positive (red) probe of the DMM to the positive (+) terminal of the battery.
4. Connect the negative (black) probe of the DMM to the negative (-) terminal of the battery.
5. Read the voltage displayed on the DMM.

A fully charged battery should show a voltage of around 12.6 volts.

If the engine is running, the charging voltage should be between 13.7 and 14.7 volts.

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Two charges one positive and one negative both with a charge of 1.1X10-10 C. They are 10-6 m apart. A third charge which is positive is located half between the first 2 charges, and the third charge is 10-17 C. What is the magnitude of force on the third charge

Answers

The magnitude of the force on the third charge is 1.21 x [tex]10^-^5[/tex] N, acting along the line between charges.

To calculate the magnitude of the force on the third charge, we can use Coulomb's Law:

F = k * (q1 * q2) / [tex]r^2[/tex], where

F is the force,

k is the electrostatic constant (8.99 x [tex]10^9[/tex] N [tex]m^2[/tex]/[tex]C^2[/tex]),

q1 and q2 are the charges, and

r is the distance between them.

The third charge is equidistant to both first and second charges.

Therefore, calculate the force between the third charge and each of the other charges separately and then add them vectorially.

The forces from each charge are equal in magnitude, 1.21 x[tex]10^-^5[/tex] N, and act along the line between charges.

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A baseball of mass 146 g is thrown with a velocity of < 23, 23, -14 > m/s. What is the kinetic energy of the baseball

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The kinetic energy of the baseball is approximately 162.18 J (joules).

To calculate the kinetic energy of the baseball, we use the formula:

Kinetic Energy (KE) = 0.5 * mass * velocity²

First, we need to convert the mass of the baseball from grams to kilograms:

146 g = 0.146 kg

Next, we need to calculate the magnitude of the velocity vector:

|velocity| = √(23² + 23² + (-14)²) = √(529 + 529 + 196) = √1254 ≈ 35.41 m/s

Now, we can calculate the kinetic energy:

KE = 0.5 * 0.146 kg * (35.41 m/s)² ≈ 162.18 J

The kinetic energy of an object is the energy it possesses due to its motion. It depends on both the mass and the velocity of the object. In this case, we have a baseball with a mass of 146 g and a given velocity vector. To find the kinetic energy, we first converted the mass to kilograms, then calculated the magnitude of the velocity vector, and finally used the kinetic energy formula to find the answer. The kinetic energy of the baseball is approximately 162.18 J.

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Moment of inertia of an area about an axis Question 4 options: Should be always zero if the axis is centroidal axis Can be zero, greater than zero or less than zero, it depends on the area

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The main answer to the question is that the moment of inertia of an area about an axis can be zero, greater than zero, or less than zero, depending on the area. This is because the moment of inertia is a measure of an object's resistance to rotational motion and is affected by both the shape and orientation of the area.



If the axis of rotation is the centroidal axis, then the moment of inertia will be minimized and can potentially be zero.

However, if the axis is not located at the centroid, the moment of inertia can vary greatly.



In summary, the moment of inertia of an area about an axis can be zero, greater than zero, or less than zero, and is dependent on the shape and orientation of the area as well as the location of the axis of rotation.

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. The wavelengths of visible light range from approximately 400 to 750 nm. What is the corresponding range of photon energies for visible light

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The corresponding range of photon energies for visible light is approximately [tex]2.65 * 10^{-19} J to 4.96 * 10^{-19} J[/tex].

The energy of a photon is directly proportional to its frequency or inversely proportional to its wavelength. Therefore, the range of photon energies for visible light can be calculated using the equation E=hc/λ, where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of the light.
Substituting the minimum and maximum wavelengths of visible light, we get:
E_min = hc/λ_max = [tex](6.626 * 10^{-34} Js)(3.0 * 10^8 m/s)/(750 * 10^{-9} m) = 2.65 * 10^{-19} J[/tex]
E_max = hc/λ_min = [tex](6.626 * 10^{-34} Js)(3.0 * 10^8 m/s)/(400 * 10^{-9} m) = 4.96 * 10^{-19} J[/tex]

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