mars has two moons, phobos and deimos, that are most similar in character to ______________

(a) The Jeans mass (M_J) is a critical mass above which a cloud in the interstellar medium becomes gravitationally unstable and begins to collapse. It can be calculated using the formula: M_J = (5 * k * T * (3/2) / (G * μ * m_H))^(3/2) * (1/n)^(1/2)

where k is the Boltzmann constant (1.38 x 10^(-23) J/K), T is the temperature, G is the gravitational constant (6.67 x 10^(-11) N m²/kg²), μ is the mean molecular weight, m_H is the mass of a hydrogen atom (1.67 x 10^(-27) kg), and n is the number density.
For the molecular (H2) cloud: T = 10 K, n = 10^12 m^(-3), and μ = 2 (since H2 has two hydrogen atoms).
M_J(H2) ≈ 1.26 x 10^(-17) kg
For the atomic (H) cloud: T = 120 K, n = 10^7 m^(-3), and μ = 1 (since H has one hydrogen atom).
M_J(H) ≈ 3.36 x 10^22 kg
(b) The minimum radius (R_J) for each cloud to collapse can be found using the formula:
R_J = (3 * M_J / (4 * π * ρ))^(1/3)
where ρ = n * μ * m_H is the mass density of the cloud.
R_J(H2) ≈ 1.73 x 10^6 m
R_J(H) ≈ 2.31 x 10^(-2) m
(c) The timescale (t_collapse) for the gravitational collapse of each cloud can be estimated using the formula:
t_collapse ≈ (3 * π / (32 * G * ρ))^(1/2)
t_collapse(H2) ≈ 4.54 x 10^11 s ≈ 14,400 years
t_collapse(H) ≈ 1.02 x 10^16 s ≈ 3.23 x 10^8 years

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The temperature change if the system is a diatomic gas is 0.046K.

For a monatomic gas, the temperature change ΔT is given by ΔT = ΔE / (3/2 R n), where ΔE is the change in thermal energy, R is the gas constant, and n is the number of moles. Plugging in the given values, we get ΔT = (1.00 J) / (3/2 * 8.314 J/(mol*K) * 0.900 mol) ≈ 0.077 K.

For a diatomic gas, the temperature change ΔT is given by ΔT = ΔE / (5/2 R n), where ΔE, R, and n are the same as in the monatomic case. Plugging in the given values, we get ΔT = (1.00 J) / (5/2 * 8.314 J/(mol*K) * 0.900 mol) ≈ 0.046 K.

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The hare has the greater instantaneous speed since he can go faster.

In a race, both the turtle and the hare travel the same distance. The tortoise travels gradually and finishes the race before the hare, whereas the hare travels very quickly for brief periods of time but frequently stops.

Since he can move more quickly, the hare has a faster instantaneous speed. Since the turtle won, he moves at a faster average speed.

An object's instantaneous speed is a measure of how quickly it is travelling right now. The speed and direction of an object's motion are both included in its instantaneous velocity, which is a vector quantity.

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Answer:

Explanation:

35,000 Joules of work is done by the external force in pulling the 350-kg box 7 meters up the 30-degree frictionless incline plane.

Given the information provided, a 350-kg box is pulled 7 meters up a 30-degree incline plane by an external force of 5000 N acting parallel to the frictionless plane. The work done by the external force can be calculated using the formula: Work = Force x Distance x cos(theta), where theta is the angle between the force and the displacement.

In this case, since the external force is parallel to the incline plane, the angle (theta) is 0 degrees. Therefore, the work done is:

Work = 5000 N x 7 m x cos(0 degrees)
Work = 5000 N x 7 m x 1
Work = 35,000 J (Joules)

So, 35,000 Joules of work is done by the external force in pulling the 350-kg box 7 meters up the 30-degree frictionless incline plane.

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The ratio of the width of the 51.5-m-wide opening to the 19.1 m wavelength of the ocean waves approaching it straight on is approximately 2.7. This information can be useful in further analysis of wave behavior and potential effects on the harbor entrance.

To analyze this situation, we can consider the ratio of the width of the opening to the wavelength of the ocean waves.

Calculate the ratio of the width of the opening to the wavelength of the waves.
Ratio = (Width of the opening) / (Wavelength of the waves)

Plug in the given values.
Ratio = (51.5 m) / (19.1 m)

Calculate the result.
Ratio ≈ 2.7

So, the ratio of the width of the 51.5-m-wide opening to the 19.1 m wavelength of the ocean waves approaching it straight on is approximately 2.7. This information can be useful in further analysis of wave behavior and potential effects on the harbor entrance.

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The transit method gives us the radius of the exoplanet if we know the star's properties. The correct answer is option B.

The transit method is one of the most commonly used techniques for detecting exoplanets. This method involves observing a star and looking for periodic dips in its brightness, which occur when an exoplanet passes in front of the star from our line of sight. By carefully measuring the shape and depth of these dips, astronomers can determine the size or radius of the exoplanet relative to the star. This is possible because the amount of light that is blocked by the planet during transit is directly related to the size of the planet relative to the size of the star.

However, to determine other properties such as the mass and density of the exoplanet, additional measurements are typically needed, such as radial velocity measurements or modeling the planet's atmospheric composition and structure.

Therefore option B is the correct answer.

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The Hubble Space Telescope obtains higher-resolution images than most ground-based telescopes because it is positioned above the Earth's atmosphere.

The atmosphere can cause distortion and blurring of images, but the Hubble's location allows it to capture clear and detailed images of the universe. Additionally, the Hubble has a larger mirror than many ground-based telescopes, allowing it to collect more light and produce sharper images.

A telescope in space used to observe celestial objects is known as a space telescope or space observatory. The American Orbiting Astronomical Observatory, OAO-2 launched in 1968, and the Soviet Orion 1 ultraviolet telescope aboard space station Salyut 1 in 1971 were the first operational telescopes, both of which were proposed by Lyman Spitzer in 1946. Space telescopes do not experience the light pollution that ground-based observatories do, nor do they experience the filtering and distortion (scintillation) of the electromagnetic radiation they detect.

The two main categories of satellites are those that map the entire sky (astronomical surveys) and those that concentrate on certain astronomical objects or regions of the sky and beyond. Space telescopes are separate from Earth imaging satellites, which are used for weather study, espionage, and other sorts of satellite imaging that point towards the Earth.

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The primary stress that plays a role in keeping the mount securely attached to the wall is shear stress.

Shear stress is the stress that acts parallel to the surface of the material. This is because the mounting plate is flush with the wall and the weight of the mirror exerts a force parallel to the wall, causing the plate to experience shear stress.

However, it is important to note that other stresses such as tension stress and compression stress may also come into play depending on the specific design of the mount and the materials used.

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In this case the reference temperature for the calculation of the enthalpy of carbon is 0°F.

To find the reference temperature for the calculation of the enthalpy of carbon, we need to use the equation for enthalpy, which is given by H = ∫Cp dT.

We know the value of enthalpy at 1000°F, which is 1510 Btu/lb. Using this value and the equation for Cp given in the problem, we can integrate to find the reference temperature.

First, we need to express Cp as a function of temperature in the integral.

We can do this by rearranging the equation:

Cp = 1.2 + 0.0050 T - 0.0000021 T2 = 1.2T0 + 0.0050T1 - 0.0000021T2 where T0 = 1, T1 = T, and T2 = T².

Now, we can integrate to find the enthalpy:

H = ∫Cp dT = 1.2T + 0.0050/2 T² - 0.0000021/3 T³ + C

where C is the constant of integration.

We can determine the value of C by using the given enthalpy value at 1000°F:

1510 = 1.2(1000) + 0.0050/2 (1000)² - 0.0000021/3 (1000)³ + C

Solving for C, we get:

C = 1510 - 1200 - 250 + 0.7 = 60.7

Substituting this value of C back into the integral equation, we get:

H = 1.2T + 0.0050/2 T² - 0.0000021/3 T³ + 60.7

Now, we can use the fact that enthalpy is defined relative to a reference temperature of 0°F.

So, we need to set T = 0 in the equation and solve for the constant term:

H(0) = 1.2(0) + 0.0050/2 (0)² - 0.0000021/3 (0)³ + 60.7 = 60.7

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The problem of determining the mass of a jet launched from an aircraft carrier catapult and the power output of the catapult involves the field of mechanics. The problem requires us to calculate the mass of the jet given the work done by the catapult, as well as the power output of the catapult given the time it takes to launch the jet.To solve this problem, we can use the principles of work and energy, which are defined as the product of force and displacement, and the ability to do work, respectively. We can use the given work done by the catapult to determine the change in kinetic energy of the jet, and then use this information to calculate the mass of the jet.To determine the power output of the catapult, we can use the relationship between work and power, which is defined as the rate at which work is done. By dividing the work done by the catapult by the time it takes to launch the jet, we can determine the power output of the catapult.Overall, this problem demonstrates the application of mechanics principles to solve a real-world problem involving the motion of objects and the forces that act on them. By understanding the principles of work, energy, and power, we can analyze and optimize systems for a wide range of applications in industry, transportation, and other fields.

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The power output of the catapult is 1.4x[tex]10^7[/tex] watts.

To find the mass of the jet, we can use the kinetic energy equation:

KE = [tex]1/2 mv^2[/tex]

where KE is the kinetic energy, m is the mass, and v is the final velocity.

We know the final velocity (67 m/s) and the work done (7.0x[tex]10^7 J[/tex]). The work done is equal to the change in kinetic energy:

Work = KE_final - KE_initial

Since the jet starts from rest, the initial kinetic energy is zero:

Work = KE_final - 0

KE_final = Work

KE_final =[tex]7.0x10^7 J[/tex]

Now we can use the kinetic energy equation to find the mass:

KE = 1/2 [tex]mv^2[/tex]

7.0x[tex]10^7 J[/tex] = 1/2 m (67[tex]m/s)^2[/tex]

m = 1.0x[tex]10^5 kg[/tex]

So the mass of the jet is 1.0x[tex]10^5 kg.[/tex]

To find the power output of the catapult, we can use the work-energy theorem:

Work = Power x Time

We know the work done (7.0x[tex]10^7 J[/tex]) and the time (5.0 s), so we can solve for power:

Power = Work / Time

Power = 7.0x[tex]10^7 J[/tex]/ 5.0 s

Power = 1.4x10^7 W

So the power output of the catapult is 1.4x[tex]10^7[/tex] watts.

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The kinetic energy of a 30-gram bullet that is traveling at 200 m/s can be calculated using the formula KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity. Plugging in the values, we get:
KE = (1/2) * 0.03 kg * (200 m/s)^2
KE = 600 J

Therefore, the kinetic energy of the 30-gram bullet traveling at 200 m/s is 600 Joules.
The kinetic energy of a 30-gram bullet traveling at 200 m/s can be calculated using the formula:
Kinetic Energy (KE) = 0.5 * mass * velocity^2

First, convert the mass of the bullet to kilograms (1 gram = 0.001 kg):
30 grams * 0.001 = 0.03 kg
Next, plug in the values into the formula:
KE = 0.5 * 0.03 kg * (200 m/s)^2
KE = 0.5 * 0.03 kg * 40,000 m^2/s^2
KE = 600 Joules

So, the kinetic energy of the 30-gram bullet traveling at 200 m/s is 600 Joules.

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The brute-force magnetic field along the z-axis is zero. The mechanical energy of the magnetic moment as a function of time is constant.

Since the circular loop lies on the xy-plane, the magnetic field it produces is also confined to the xy-plane. Therefore, the brute-force magnetic field along the z-axis is zero.

As the magnetic moment flies through the loop, it experiences a changing magnetic field due to the motion of the magnetic moment relative to the loop.

This changing magnetic field induces an electric field, which in turn generates a magnetic force on the magnetic moment. However, since the magnetic moment is flying directly along the z-axis, it does not experience any magnetic force.

Therefore, the mechanical energy of the magnetic moment is simply the sum of its kinetic and potential energies, which is constant since there are no external forces acting on the magnetic moment. This means that the mechanical energy as a function of time is a horizontal line.

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The rate at which a capacitor can store charge on its plates is known as its "capacitance." Capacitance is defined as the ratio of the electric charge stored on each plate of a capacitor to the potential difference between the plates.

It is measured in units of Farads (F), named after the British scientist Michael Faraday. The capacitance of a capacitor is determined by factors such as the surface area of the plates, the distance between them, and the type of dielectric material between them.

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The potential energy of the two charges when they are 24 nm apart is 1.25 J.

According to the given information, the potential energy of two electric charges is inversely proportional to the distance between them. This means that as the distance between the charges decreases, their potential energy will increase.

Using the formula for potential energy, we can set up the following equation:

1.0 J = k(Q1Q2)/d

where k is the Coulomb's constant, Q1 and Q2 are the charges, and d is the distance between them.

We can rearrange the equation to solve for k:

k = (1.0 J * d) / (Q1Q2)

Now that we have the value of k, we can use it to find the potential energy when the distance between the charges is 24 nm:

PE = k(Q1Q2)/d

PE = k(Q1Q2)/24 nm

PE = [(1.0 J * 30 nm) / (Q1Q2)] * (Q1Q2) / 24 nm

PE = 1.25 J

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When Elena returns to Earth, she will be 17.5 years/ 18 year option (4)

According to special relativity, time dilation occurs for a moving observer relative to a stationary observer. In this scenario, Elena is the moving observer while Olga is stationary on Earth. We need to find out the time experienced by Elena during her journey, which is different from the time experienced by Olga.

From Elena's perspective, the distance travelled is given by:

d = v*t = (0.6c)5 years = 0.6299792458 m/s * 5 years ≈ 895,800,000 km

Since Elena travels at a high velocity, time dilation occurs. The time experienced by Elena is given by:

t' = t / sqrt(1 - (v/c)^2)

where t is the time experienced by Olga on Earth, v is the velocity of Elena relative to Olga, and c is the speed of light.

For the outbound journey, we have:

t' = 5 / sqrt(1 - (0.6c/c)^2) = 5 / 0.8 = 6.25 years

For the return journey, the time experienced by Elena is also 6.25 years. Thus, the total time experienced by Elena during her journey is:

t_total = 5 + 6.25 + 6.25 = 17.5 years

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The acceleration of the cart can be expressed using Newton's second law as a = (F - f) / m. Given the cart's mass is 3.0 kg, the applied force is 12.6 N, and the friction force is 5.4 N, the cart's acceleration is 2.4 m/s².

Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, it can be expressed as F = ma, where F is the net force acting on the object, m is its mass, and a is its acceleration.

In this problem, the net force acting on the cart is given by F - f, where F is the applied force and f is the friction force. Therefore, using Newton's second law, we can write: F - f = ma

Substituting the given values, we get:

12.6 N - 5.4 N = 3.0 kg x a

Simplifying, we get:

7.2 N = 3.0 kg x a

Dividing both sides by 3.0 kg, we get:

a = 7.2 N / 3.0 kg

a = 2.4 m/s²

Therefore, the cart's acceleration is 2.4 m/s².

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The average magnetic field strength of the magnet is approximately 2.04545 T.

When a conductor carrying current is placed in a magnetic field, the conductor experiences a magnetic force, F, given by the formula

                    F = B * I * L * sinФ

where F is the force experienced by the wire (2.25 N), I is the current passing through the wire (22.0 A), and L is the length of the wire in the magnetic field (5.00 cm, which we need to convert to meters), Ф is the angle of wire with the magnetic field, here Ф = 90°

sinФ= sin 90°

       = 1

so, F = B * I * L
The direction of force is always perpendicular to the plane containing both the conductor and the magnetic field given by Flemig's left-hand rule.

First, let's convert the length of the wire to meters:
5.00 cm = 0.0500 m

Now, we can rearrange the formula to solve for B, the magnetic field strength:
            B = F / (I * L)

Plug in the given values:
             B = 2.25 N / (22.0 A * 0.0500 m)

Perform the calculation:
B ≈ 2.04545 T

So, the average magnetic field strength of the magnet is approximately 2.04545 T.

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Using the improved Euler method with step sizes 0.5 and 0.25, we get 35.63% and 39.95% of the limiting velocity after 1 and 2 seconds respectively.

In this issue, we are given an underlying worth issue to find the descending speed of an item as it falls in the wake of opening a parachute. We will utilize the superior Euler technique to estimated the arrangement with step sizes of 0.5 and 0.25.

To apply the better Euler strategy, we utilize the emphasis equation, which includes ascertaining the subsidiary of the capability at each step. For this situation, the subordinate is given by f(t, v) = - g - (k/m) * v^2, where g is the gravitational speed increase, k is the drag coefficient, m is the mass of the item, and v is the speed.

Utilizing the underlying condition v(0) = 0, we can utilize the better Euler strategy to rough the answer for v(1) and v(2) with step sizes h = 0.5 and h = 0.25. With h = 0.5, we get v(1) = - 22.12 m/s and v(2) = - 36.38 m/s. With h = 0.25, we get v(1) = - 29.17 m/s and v(2) = - 45.47 m/s.

The restricting speed of the item, otherwise called the maximum speed, is given by - sqrt(gm/k) = - 58.78 m/s. To find the level of the restricting speed achieved following 1 second and 2 seconds, we basically partition the determined speeds by the restricting speed and increase by 100.

Accordingly, following 1 second, the level of the restricting speed accomplished is 37.6% with h = 0.5 and 49.5% with h = 0.25. Following 2 seconds, the level of the restricting speed accomplished is 61.7% with h = 0.5 and 77.3% with h = 0.25. These outcomes show that the article's speed moves toward the restricting speed as it falls, and the pace of approach increments over the long haul.

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The complete question is:

Falling parachutist As in Problem 25 of Section 2.4, you bail out of a helicopter and immediately open your parachute, so your downward velocity satisfies the initial value problem dv dt = 32 – 1.6v, v(0) = 0 (with t in seconds and v in ft/s). Use the improved Euler method with a programmable calculator or computer to approximate the solution for 0 31 32, first with step size h = 0.01 and then with h = 0.005, rounding off approx- imate v-values to three decimal places. What percentage of the limiting velocity 20 ft/s has been attained after 1 second? After 2 seconds?

The velocity of a river depends on its depth, width, and discharge. In this scenario, we know that both streams have the same depth and discharge, but stream b is half as wide as stream a.

Therefore, the velocity of stream b must be greater than that of stream a. This is because the same volume of water has to flow through a narrower channel in stream b, which means that the water has to move faster to maintain the same discharge rate.

This is similar to a garden hose: if you partially cover the opening of the hose with your finger, the water comes out faster because it has to flow through a smaller opening. So, in conclusion.

even though both streams have the same depth and discharge, stream b has a greater velocity because it is narrower than stream a.

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The steady-state creep rate at 1250 K and a stress level of 50 MPa for the given stainless steel is 1.25 x 10^-12 s^-1.

To compute the steady-state creep rate at 1250 K and a stress level of 50 MPa for the given stainless steel, we can use the equation for steady-state creep rate:

[tex]\epsilon ss = A\sigma ^nexp(-Q/RT)[/tex]

First, we need to find the value of A :

[tex]ln(\epsilon ss) = ln(A) + n ln(\sigma) - Q/RT[/tex]

Taking the natural logarithm of both sides of equation (5-1) :

[tex]ln(\epsilon s) = -12.09 + 7.0 ln(70)[/tex]

[tex]ln(2.5 * 10^-3) = -8.29 + 7.0 ln(70)[/tex]

Solving these equations simultaneously for ln(A) and Q/RT, we get:

ln(A) = -24.36

Q/RT = 22,832 K

Next, we can use the values of A and Q/RT :

[tex]\epsilon ss = A\sigma ^nexp(-Q/RT)[/tex]

[tex]\epsilon ss = 10^{(-24.36)} * (50)^7 * exp(-22,832/1250R)[/tex]

where R is the gas constant in J/(mol·K) = 8.314 J/(mol·K)

[tex]\epsilon ss = 1.25 * 10^-12 s^-1[/tex]

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(A) Yes, the bee can detect the difference between flowers. (B) The magnitude of the difference between the electric fields of the flower is

To calculate the magnitude of the difference between the electric fields of the flowers, we need to use the formula for the electric field:

E = kQ/r²

where k is Coulomb's constant (9 x 10⁹ N m²/C²), Q is the charge on the flower, and r is the distance between the bee and the flower.

For the flower with a charge of +30 pC:

E1 = (9 x 10⁹ N m²/C²) x (30 x 10⁻¹² C) / (0.24 m)²
E1 = 4.69 N/C

For the flower with a charge of +40 pC:

E2 = (9 x 10⁹ N m²/C²) x (40 x 10⁻¹² C) / (0.24 m)²
E2 = 6.25 N/C

The magnitude of the difference between the fields of the flowers is:

|AE| = |E2 - E1|
|AE| = |6.25 N/C - 4.69 N/C|
|AE| = 1.56 N/C

Therefore, the bee can detect the difference between the electric fields of the flowers, since the difference is larger than the smallest change in the field that honeybees can detect (0.77 N/C).

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Standard conditions refer to a specific set of temperature, concentration, and pressure values, which are commonly used in scientific experiments and calculations to provide a consistent basis for comparison.

In most thermodynamic calculations, standard conditions are defined as a temperature of 298.15 K (25°C), a concentration of 1 mol/L (1 M) for solutions, and a pressure of 1 atm (101.325 kPa) for gases. These conditions are important because they serve as a reference point, allowing scientists to easily compare the properties and behavior of various substances under the same set of conditions.

The use of standard conditions simplifies calculations and ensures that the results are consistent across different studies. By providing a uniform reference point, researchers can focus on the effects of specific variables, such as the type of substance, its structure, or its interactions with other substances, without having to account for variations in temperature, concentration, or pressure.

To summarize, standard conditions for most thermodynamic calculations involve a temperature of 298.15 K (25°C), a concentration of 1 mol/L (1 M) for solutions, and a pressure of 1 atm (101.325 kPa) for gases. These conditions provide a consistent basis for comparison, enabling scientists to examine the properties and behavior of various substances under the same set of conditions.

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The final parameters of the problem are the conservation of charge and the definition of capacitance.

According to the conservation of charge, the total charge in the system before and after the capacitors are connected in parallel must be the same. Therefore, the total charge stored by the capacitors is the same before and after the connection.

According to the definition of capacitance, the capacitance of the equivalent parallel combination of the capacitors is the sum of the individual capacitances. Therefore, the final capacitance of the parallel combination can be calculated as the sum of the capacitances of the individual capacitors.

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Cloud-to-cloud lightning makes up 60 % of all lightning.

Cloud-to-cloud lightning is a type of lightning that occurs within a single thunderstorm cloud, between two different thunderstorm clouds or between a thunderstorm cloud and the air near the ground. It is also known as intra-cloud lightning.

This type of lightning is responsible for creating the beautiful light shows that we often see during thunderstorms, but it can also be dangerous and cause damage to electrical equipment.
In terms of its frequency compared to other types of lightning, cloud-to-cloud lightning makes up a significant portion of all lightning strikes.

However, it is difficult to determine an exact percentage because lightning is a complex and unpredictable phenomenon. Studies have shown that cloud-to-cloud lightning accounts for around 60% of all lightning strikes worldwide.
This high percentage can be attributed to the fact that thunderstorm clouds often contain multiple charged regions, which can create the conditions necessary for cloud-to-cloud lightning.

While cloud-to-cloud lightning may not be the most well-known type of lightning, it plays an important role in the overall frequency and impact of lightning strikes worldwide.

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When the incident wave vector and the diffracted wave vector satisfy the Bragg condition, they will intersect with a reciprocal lattice point on the Ewald sphere.

To depict the satisfaction of the Bragg condition for the (020) planes in the FCC crystal, we can use an Ewald sphere-reciprocal lattice construction. Here are the steps:

1. Draw reciprocal lattice points along the y direction.
2. Place the tip of the incident wave vector on the origin of the reciprocal lattice and draw the incident wave vector back to the origin of the Ewald sphere.
3. Draw the diffracted wave vector from the origin of the Ewald sphere to the 020 reciprocal lattice point incident and diffracted wave vectors.
4. Draw the radius of the Ewald sphere through the ends of the incident and diffracted wave vectors.

The reciprocal lattice represents the Fourier transform of the direct lattice, which allows us to describe the diffraction pattern of a crystal. The Ewald sphere represents the energy conservation and momentum conservation in the diffraction process. The intersection of the Ewald sphere with the reciprocal lattice points represents the allowed diffraction orders that satisfy the Bragg condition.

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The given problem involves calculating the force between two parallel wires carrying a current, given their length, separation distance, and current. Specifically, we are asked to determine the new force between the wires when the separation distance is doubled.To calculate the force between the wires, we need to use the formula for the force between two parallel wires carrying a current.

The formula is given by F = μ0*I1*I2*L/d, where F is the force, μ0 is the permeability of free space, I1 and I2 are the currents in the wires, L is the length of the wires, and d is the separation distance between the wires.In this case, we are given the length of the wires, the separation distance, and the current in the wires. We can use these values to calculate the force between the wires using the formula.To determine the new force when the separation distance is doubled, we can use the formula again with the new separation distance and the same values for the other variables.The final answer is a number, which represents the new force between the wires in Newtons.

Overall, the problem involves applying the principles of electricity and force to determine the force between two parallel wires carrying a current, given their length and separation distance. It also requires an understanding of the formula for the force between two parallel wires and how it depends on the variables involved.

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The required diameter of the beam at the fixed end is c. 292 mm.

Why the required diameter of the beam at the fixed end is c. 292 mm?

To find the required diameter of the beam at the fixed end, we can use the formula for maximum bending induced shear stress in a cantilever beam:

τ_max = (3/2) ˣ (V/A)

where τ_max is the maximum bending induced shear stress, V is the maximum shear force at the fixed end, A is the cross-sectional area of the beam at the fixed end.

First, we need to find the maximum shear force at the fixed end of the cantilever beam. This can be done using the formula for the total load on the beam:

W = wL

where W is the total load, w is the uniformly distributed load, and L is the span of the beam. Substituting the given values, we get:

W = 30 kN/m ˣ 2 m = 60 kN

Next, we can find the maximum shear force at the fixed end using the formula:

V = W/2

Substituting the value of W, we get:

V = 60 kN/2 = 30 kN

Now we can use the formula for maximum bending induced shear stress to find the required diameter of the beam at the fixed end. Rearranging the formula, we get:

A = (3/2) ˣ (V/τ_max)

Substituting the given values, we get:

A = (3/2)ˣ (30 kN/(2 N/mm²)) = 67,500 mm²

Finally, we can use the formula for the area of a circular cross section to find the required diameter of the beam:

A = (π/4) ˣ D²

Rearranging the formula, we get:

D = sqrt((4ˣA)/π)

Substituting the value of A, we get:

D = sqrt((4ˣ67,500 mm²)/π) = 292 mm

Therefore, the required diameter of the beam at the fixed end is c. 292 mm.

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NGC 1300 is a tilted disk galaxy with flocculent discontinuous spiral arms. It does not have a bar, and its bulge is not very large. NGC 4414 is an edge-on spiral with a large bulge. Therefore, it is an Sa galaxy. M101 is obviously a barred spiral.

Hubble's galaxy classification system categorizes galaxies based on their visual appearance. It consists of three main categories: elliptical galaxies, spiral galaxies, and irregular galaxies. Spiral galaxies are further divided into subcategories based on the size of their bulge and the tightness of their spiral arms.

NGC 1300 is a tilted disk galaxy with flocculent discontinuous spiral arms, which indicates that it does not have a well-defined structure. It is classified as either an Sc or Sb galaxy. NGC 4414 is an edge-on spiral with a large bulge and tightly wrapped arms, which makes it an Sa galaxy. Finally, M101 is an obviously barred spiral galaxy with tightly wrapped arms, making it an SBb or SBe galaxy.

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--The complete question is, How would you classify the listed galaxies using the system of Hubble's galaxy classes?

Drag the items on the left to the appropriate blanks on the right.

_____________ is a tilted ditik galaxy with a flocculent discontinuous spiralarms. It does not have a bar, and its bulge is not very large It should be Sc or Sb galaxy.

_________  is an edge-on spiral with a large bulge. It does not show the bar and its arms are tightly wrapped, therefore it is an Sa galaxy.

_________________ is obviously a barred spiral. It is an SBb or SBe galaxy given how tightly its spiral arms are wrapped.

Options:

NGC 1300

NGC 4414

M101--

A sound wave that is travelling at [tex]343\frac{m}{s}[/tex] that is emitted by the foghorn of a tugboat, with an echo heard [tex]2.30 s[/tex], then the reflecting object is 394.45 meters away from the tugboat.

To determine how far away the reflecting object is when a sound wave is travelling at [tex]343\frac{m}{s}[/tex] and an echo is heard 2.30 s later, you should follow these steps:

1. Calculate the total distance the sound wave travels using the speed of sound and the time taken for the echo to return. Since the speed of sound is [tex]343\frac{m}{s}[/tex]  and the time is 2.30 s, multiply these values together:
  Total distance = [tex]343\frac{m}{s}\times2.30 s = 788.9 m[/tex]

2. Since the sound wave travels to the reflecting object and then back to the source, divide the total distance by 2 to find the distance to the reflecting object:
  Distance to the reflecting object [tex]= \frac{{788.9 \ }}{2}m = 394.45m[/tex]

So, the reflecting object is 394.45 meters away from the tugboat.

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