Which statements accurately compare the tectonic activity of the planets? Check all that apply. The terrestrial planets experience quakes; the gas giants do not. The gas giants have multiple plates; the terrestrial planets do not. The gas giants have no true surfaces, so they experience tectonic activity. Terrestrial planets experience (or experienced) tectonic activity; gas giants do (or did) not. Mountains and volcanoes are found on the gas giants, not on terrestrial planets. Activity in the molten interior of the terrestrial planets results (or resulted) in tectonic activity.

The electron configuration of an atom describes the arrangement of its electrons in various energy levels and orbitals. The total number of electrons in an atom is equal to its atomic number.

(a) Boron has an atomic number of 5, meaning it has 5 electrons. The electron configuration of boron is 1s²2s²2p¹, that is it has 2 electrons in the 1s orbital, 2 electrons in the 2s orbital, and 1 electron in the 2p orbital. The total number of electrons is 2+2+1=5.

(b) Aluminum has 13 electrons since it has an atomic number of 13. Aluminum's electron configuration is 1s²2s²2p⁶ 3s²3p¹, which means it contains two electrons in the 1s orbital, 2 electrons in the 2s orbital, 6electrons in the 2p orbital, 2 electrons in the 3s orbital, and 1 electron in the 3p orbital. The total number of electrons is correct: 2+2+6+2+1=13.

(c) Potassium has an atomic number of 19, meaning it has 19 electrons. The electron configuration of potassium is 1s²2s²2p⁶ 3s²3p⁶ 4s¹, that is it has 2 electrons in the 1s orbital, 2 electrons in the 2s orbital, 6 electrons in the 2p orbital, 2 electrons in the 3s orbital, 6 electrons in the 3p orbital, and 1 electron in the 4s orbital. Total number of electrons is 2+2+6+2+6+1=19.

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The angle of the red line (λ = 656 nm) in the ∅₁ 1st order is 0.0321 radians.

An optical lattice is a periodic structure formed by interfering laser beams, which creates a spatially varying pattern of light and dark regions. The lattice serves as a diffraction grating that can split light into its constituent wavelengths, enabling scientists to study the properties of light and matter. Optical lattices are widely used in experiments involving cold atoms, quantum optics, and condensed matter physics.

The angle of the red line (λ = 656 nm) in the ∅₁ 1st order can be calculated using the formula:

sin(θ) = mλ/d

where θ is the angle between the incident light and the diffracted light, m is the order of the diffraction (m = 1 for the ∅₁ 1st order), λ is the wavelength of the light, and d is the spacing between the diffracting elements of the grating (d = 1/560 mm in this case).

Substituting the values given, we get:

sin(θ) = (1)(656 nm)/(1/560 mm)

sin(θ) = 0.0005596

Taking the inverse sine of both sides, we get:

θ = sin⁻¹(0.0005596)

θ = 0.0321 radians

Therefore, the angle of the red line (λ = 656 nm) in the ∅₁ 1st order is 0.0321 radians.

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The intensity of waves is 9.96 × 10⁻³W/m².

Power radiated, P = 50 W

Frequency of the sound, f = 1000 Hz

a) Distance, r = 20 m

The intensity of waves,

I = P/4πr²

I = 50/(4π × 20²)

I = 9.96 × 10⁻³W/m²

b) We know that, the intensity of waves,

I ∝ 1/r²

So, I/I' = (r'/r)²

Therefore,

r' = r(√I/I')

r' = 20(√9.96 × 10⁻³/1)

r' = 0.19 m

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The experiment to measure the speed of sound will involve recording the time for the loudest sound to be heard.

The dependent variable in the experiment is the time and the independent variable is the frequencies of the tuning fork.

Some potential reasons for the two lab groups in different locations coming up with different results include:

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The child's speed at the lowest point is 5.42 m/s.

At the highest point of the swing, the child is momentarily at rest and has only potential energy. At the lowest point, the child has only kinetic energy.

Using the conservation of mechanical energy, we can write:

Potential energy at highest point = Kinetic energy at lowest point

mgh = (1/2)mv²

where m is the mass of the child, g is the acceleration due to gravity, h is the height of the swing at the highest point, and v is the speed of the child at the lowest point.

First, we need to find the height of the swing at the highest point. Since the swing is supported by two chains, the height of the swing at the highest point is half the length of the chains:

h = (1/2)3.00 m = 1.50 m

Next, we can solve for the child's speed at the lowest point:

mgh = (1/2)mv²

40.0 kg * 9.81 m/s² * 1.50 m = (1/2) * 40.0 kg * v²

588 J = 20.0 kg * v²

v² = 29.4 m²/s²

v = 5.42 m/s

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The student measured the length of a wire four times and the total length from the reading is 18.55 m.

From the given,

The length of the wire measured by the student using a meter rule is :

l₁ = 18.6 m

l₂ = 18.5m

l₃ = 18.6m

l₄ = 18.5 m

The total length of the wire is obtained from the average values of length.

Total length (L) = (l₁ + l₂ + l₃ + l₄)/4

                     L  = (18.6 + 18.5 + 18.6 + 18.5) / 4

                         = 74.2 /4

                         = 18.55

Hence, the actual length of the wire = 18.6 m.

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The mass of the Ball B is 1.35 x 10⁻⁶ kg.

Gravitational Force is described by Newton's law of universal gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

The law of universal gravitation is important in many fields, including astronomy, physics, and engineering.

The gravitational force F between two objects of masses m1 and m2 separated by a distance r is given by:

F = G(m₁m₂)/r²

where G is the gravitational constant.

We can rearrange the equation to solve for the mass of Ball B:

m₂ = Fr²/Gm₁

Substituting the given values, we get:

m₂ = (8.7 x 10⁻¹⁰ N)(3.0 m)²/(6.6743 x 10¹¹ N(m^2/kg²))(30 N)

m₂ = 1.35 x 10⁻⁶ kg

Therefore, the mass of Ball B is approximately 1.35 x 10⁻⁶kg.

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The frequency of red light having wavelength 650nm in vacuum is 4.6 x 10¹⁴ Hz.

The frequency of light can be calculated using the formula:

frequency = speed of light / wavelength

In vacuum, the speed of light is approximately 3 x 10⁸ m/s.

Converting the wavelength to meters: 650 nm = 650 x 10⁻⁹ m

Plugging these values into the formula, we get:

frequency = (3 x 10⁸ m/s) / (650 x 10⁻⁹ m) = 4.6 x 10¹⁴ Hz

Therefore, the frequency of red light with a wavelength of 650nm in vacuum is 4.6 x 10¹⁴ Hz.

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The terminal velocity of the sphere is approximately 22.68 mph.

The terminal velocity of a sphere is the constant speed at which the gravitational force pulling the sphere down is balanced by the drag force pushing the sphere up. The drag force is proportional to the velocity of the sphere, and can be calculated using the following formula:

Fd = (1/2) * rho * Cd * A * v²

where Fd is the drag force, rho is the density of the fluid (air in this case), Cd is the drag coefficient (which depends on the shape of the object), A is the cross-sectional area of the object perpendicular to the direction of motion, and v is the velocity of the object.

The gravitational force pulling the sphere down is given by:

Fg = m * g

where m is the mass of the sphere and g is the acceleration due to gravity.

At terminal velocity, the drag force is equal in magnitude to the gravitational force, so:

Fd = Fg

Substituting the expressions for Fd and Fg and solving for v, we get:

v = √((2 * m * g) / (rho * Cd * A))

where A = pi * r² is the cross-sectional area of the sphere, and r is the radius of the sphere.

Plugging in the given values, we get:

v = sqrt((2 * 0.6 * 9.81) / (1.2 * 0.47 * pi * 0.3²)) ≈ 10.13 m/s

To convert this to mph, we multiply by 2.23694:

v ≈ 22.68 mph

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The pressure of the gas will also double, but the density of the gas remains the same.

option 3.

Ideal Gas Law, states that the  pressure of the gas is inversely proportional to its volume and directly proportional to its absolute temperature.

PV = nRT

Where;

When the absolute temperature of a sample of monatomic ideal gas is doubled at constant volume, the pressure of the gas will also double, but the density of the gas remains the same.

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The fitness assessments helps as, they ensure that the program is meeting the desired goals.

option C.

Fitness assessment is defined as a process of evaluating an individual's physical fitness levels and overall health.

Fitness assessment typically involves a series of tests, measurements, and evaluations that are designed to assess various components of physical fitness, such as;

Thus, fitness assessments ensure that the program is meeting the desired goals.

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Answer: A vertical column of water can be supported to a height of about 10.3 meters (33.8 feet) by standard atmospheric pressure.

- This is because the weight of the column of water is balanced by the pressure exerted by the air above it. This phenomenon is known as hydrostatic pressure, and it is the reason why water towers and barometers work.

Answer:

potential energy

Explanation:

The velocity of the ball just before it hits the ground is 14.0 m/s

Let's solve the problem using the given equation:

[tex]v^2 = u^2 + 2as[/tex]

We know that u (initial velocity) is zero, s (distance traveled) is 10 meters, and a (acceleration due to gravity) is 9.81 m/s^2. We want to find the final velocity (v) just before the ball hits the ground.

Plugging in the given values, we get:

v^2 = 0 + 2(9.81)(10)

v^2 = 196.2

Taking the square root of both sides, we get:

v = sqrt(196.2)

v = 14.0 m/s

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--The complete Question is, A ball is dropped from a height of 10 meters. What is its velocity just before it hits the ground, assuming no air resistance? (Assume that the acceleration due to gravity is 9.81 m/s^2)

Hint: You can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration due to gravity, and s is the distance traveled.--

The force required to stop the car of mass 3250 kg is -12195.12 N.

Force is the product of mass and acceleration.

To calculate the force required to stop the car, we use the formula below

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

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The ratio of their rotational kinetic energies is 4/5 or 0.8.

Let's denote the mass and radius of the cylinder and sphere as "m" and "r", respectively. At the top of the incline, both objects have only potential energy, which is then converted to kinetic energy. At the bottom of the incline, both objects have both translational and rotational kinetic energy.

For a uniform solid cylinder, the rotational inertia is[tex]1/2 * m * r^2[/tex]. For a uniform sphere, the rotational inertia is[tex]2/5 * m * r^2[/tex]. Therefore, the ratio of their rotational kinetic energies is:

(rotational kinetic energy of sphere) / (rotational kinetic energy of cylinder)

[tex]= (2/5 * m * r^2 * (v/r)^2) / (1/2 * m * r^2 * (v/r)^2)[/tex]

= (4/5)

Therefore,  rotational kinetic energy of  sphere is 80% of  rotational kinetic energy of the cylinder at  bottom of the incline.

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--The complete Question is, A uniform solid cylinder and a uniform sphere with the same mass and radius are released from rest at the top of an incline. They both roll without slipping down the incline and reach the bottom with the same translational speed. What is the ratio of their rotational kinetic energies at the bottom of the incline?--

(a) The new angular speed of the student is 2.34 rad/s, and (b) The kinetic energy of the student before the objects are pulled in is 8.22 J, and the kinetic energy of the student after the objects are pulled in is  24.8 J.

Angular speed, also known as rotational speed, is the measure of how fast an object rotates or revolves around a central point or axis, and it is measured in radians per second (rad/s) or degrees per second (°/s). It is a scalar quantity that describes the magnitude of the rotational velocity of an object.

(a) To solve for the new angular speed of the student, we can use the conservation of angular momentum. Initially, the student, stool, and objects have an angular momentum given by:

L = Iω

where I is the moment of inertia and ω is the angular speed. Since there is no external torque acting on the system, the angular momentum is conserved. Therefore, we can write:

I1ω1 = I2ω2

where I1 and ω1 are the initial moment of inertia and angular speed, and I2 and ω2 are the final moment of inertia and angular speed.

At the initial state, the moment of inertia of the student plus stool and objects is given by:

I1 = I_student + I_objects

where I_student is the moment of inertia of the student and stool and I_objects is the moment of inertia of the objects. The moment of inertia of the objects can be approximated as:

I_objects ≈ 2mr²

where m is the mass of one object and r is the distance from the axis of rotation to the object. Substituting the given values, we get:

I1 = 3.0 kg·m² + 2(3.6 kg)(1.0 m)² ≈ 29.04 kg·m²

At the final state, the moment of inertia is given by:

I2 = I_student + 2mr²

where r is the new distance from the axis of rotation to the objects. Substituting the given values, we get:

I2 = 3.0 kg·m² + 2(3.6 kg)(0.43 m)² ≈ 8.97 kg·m²

Substituting the known values and solving for ω2, we get:

I1ω1 = I2ω2

(29.04 kg·m²)(0.75 rad/s) = (8.97 kg·m²)ω2

ω2 ≈ 2.34 rad/s

So, the new angular speed of the student is approximately 2.34 rad/s.

(b) To solve for the kinetic energy of the student before and after the objects are pulled in, we can use the formula:

KE = (1/2)Iω²

where KE is the kinetic energy, I is the moment of inertia, and ω is the angular speed.

At the initial state, the kinetic energy is given by:

KE1 = (1/2)I1ω1² ≈ 8.22 J

Substituting the known values, we get:

KE1 ≈ (1/2)(29.04 kg·m²)(0.75 rad/s)² ≈ 8.22 J

At the final state, the kinetic energy is given by:

KE2 = (1/2)I2ω2²

Substituting the known values, we get:

KE2 ≈ (1/2)(8.97 kg·m²)(2.34 rad/s)² ≈ 24.8 J

So, the kinetic energy of the student before the objects are pulled in is approximately 8.22 J, and the kinetic energy of the student after the objects are pulled in is approximately 24.8 J.

Therefore, (a) The student's new angular speed is 2.34 rad/s, and (b) the student's kinetic energy before the objects are drawn in is around 8.22 J, and the student's kinetic energy after the things are pushed in is roughly 24.8 J.

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The total resistance in the circuit is 30 ohms. Option d is correct.

In an electric circuit, resistors can be connected in different ways, such as in series or parallel. When resistors are connected in series, the total resistance is equal to the sum of the individual resistances. This is because the same current flows through each resistor, and the total voltage across the resistors is divided among them.

The resistances in series is the sum of all resistances. The three resistances are, 10 ohms, 15 ohms and 5 ohms. Therefore,

Total Resistance = 10 ohms + 15 ohms + 5 ohms = 30 ohms.

Hence, option d is correct.

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Answer: B

Explanation:

A student creates a dichotomous key to identify common household pets. The wrong  key is option  C. A pet can have a hairy body and live in a cage, or can be hairless and not live in a cage.

A dichotomous key is an instrument utilized for the classification and identification of living organisms by examining their visible traits. It is comprised of a sequence of decisions or actions which guide the user towards accurately identifying the organism.

Every decision or move involves selecting between two alternative options or features that are contradictory to each other. Furthermore, the correct answer identified as option C acknowledges that pets can possess different traits and may not necessarily conform to a specific classification.

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The final velocity of the ball 2 if the collision is perfectly elastic is 5.87 m/s.

The final velocity of the ball 2 if the collision is perfectly elastic  7.12 m/s.

The final velocity of the ball 1 if the collision is perfectly inelastic 4.4 m/s

The final velocity of the ball 2 if the collision is perfectly inelastic 4.4 m/s

Momentum is defined as product of mass and velocity of the body. It is denoted by letter p and it is expressed in kg.m/s. Mathematically p = mv. it discuss the moment of the body. body having zero mass or velocity has zero momentum. The dimensions of the momentum is [M¹ L¹ T⁻¹].

according to conservation law of momentum,

momentum before collision is equal to momentum after collision.

for elastic collision,

m₁v₁ +m₂v₂ = m₁v₁' + m₂v₂'

for inelastic collision,

m₁v₁ +m₂v₂ = (m₁+ m₂)v

where,

mass of the ball 1 m₁ = 110g = 0.11 kg

mass of the ball 2 m₂ = 340 g = 0.34kg

initial velocity of ball 1 v₁ = 15 m/s

initial velocity of ball 2 v₂ = 0 m/s

final velocity of ball 1 v₁' = ?

final velocity of ball 2 v₂' = ?

velocity of both ball v = ?

for elastic collision,

m₁v₁ +m₂v₂ = m₁v₁' + m₂v₂'

putting all the values,

0.11× 15 = 0.11v₁' + 0.34v₂'

1.65 = 0.11v₁' + 0.34v₂' _______1)

According to conservation energy in elastic collision,

1/2m₁v₁² + 1/2m₂v₂² = 1/2m₁v₁'² + 1/2m₂v₂'²

1/2×0.11×15² = 1/2×0.11×v₁'² + 1/2×0.34×v₂'²

12.37 = 0.055×v₁'² + 0.17×v₂'²______2)

Solving this simultaneous equation we get,

v₁' = 5.87 m/s

v₂'= 7.12 m/s

for inelastic collision,

m₁v₁ +m₂v₂ = (m₁+ m₂)v

0.11×18 + 0 = (0.11 + 0.34 )v

v = 4.4 m/s

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The minimum stopping distance is determined as 8 m.

The minimum stopping distance of the car is calculated as follows;

d = (u²)/(2a)

where;

when the minimum stopping distance = 2 m, initial velocity = 40 km/hr = 11.11 m/s

2 = (11.11²)/(2a)

a = (11.11²)/(2 x 2)

a = 30.86 m/s²

when the speed becomes 80 km/h, the minimum stopping distance is calculated as;

u = 80 km/h = 22.22 m/s

d = (22.22² )/ (2 x 30.86)

d = 8 m

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An astronaut measure the period of a mass spring system on Earth.

The period of a mass spring system on the moon would be longer than the period on Earth. This is because the period of a mass spring system is dependent on the square root of the ratio of the mass to the spring constant, and the acceleration due to gravity. Since the acceleration due to gravity on the moon is only 1/6th of that on Earth, the restoring force on the mass will be weaker, resulting in a longer period. Therefore, the astronaut would measure a longer period for the same mass spring system on the moon than on Earth.

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The total system momentum would remain the same before and after the collision. Option D

The law of conservation of momentum, which states that when no external forces are exerted on a system of objects, the system's overall momentum will remain constant, is a fundamental principle of physics.

The momenta of all the objects added together in the system prior to an event or interaction are equal to the momenta of the objects added together after the event or interaction.

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The velocity of the car is determined as 10.47 m/s.

The velocity of the car is calculated by applying the following formulas;

f' = f(v + vo)/(v + vs)

where:

The velocity of the car is calculated as follows;

vs = ((f' - f) / f')v

vs = ((393 - 381) / 393) x 343

vs = 10.47 m/s

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The length of the x-component of the vector is 8.1

The component that pushes right or left is called the x-component, and the part that pushes up or down is called the y-component.

To calculate the x-component of the vector plotted below, we use the formula below

Formula:

Where:

From the diagram,

Given:

Substitute these values into equation 1

Hence, the right option is B. 8.1

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Answer:

Approximately [tex]65\; {\rm g}[/tex] (assuming that the spring is horizontal, and that mass of the spring is negligible.)

Explanation:

In a simple harmonic motion, assuming that the mass is at position [tex]0[/tex] when time is [tex]0[/tex], the position [tex]x[/tex] of the mass (the oscillator) at time [tex]t[/tex] would be:

[tex]x = A\, \sin(\omega\, t)[/tex], where:

Note that [tex]\omega = 2\, \pi\, f[/tex] where [tex]f[/tex] is the frequency of the motion. It is given that [tex]f = 9.7\; {\rm s^{-1}}[/tex] in this question.

Differentiate [tex]x = A\, \sin(\omega\, t)[/tex] with respect to time [tex]t[/tex] to find the velocity of the oscillator at time [tex]t[/tex]:

[tex]v = A\, \omega \, \cos(\omega\, t)[/tex].

Differentiate again to find the acceleration of the oscillator:

[tex]a = (-A\, \omega^{2})\, \sin(\omega\, t)[/tex].

Let [tex]m[/tex] denote the mass of the oscillator and assume that mass of the spring is negligible. Assuming that the spring is horizontal, since all other force on the spring are balanced, the net force on the mass would be equal to the restoring force from the spring. By Newton's Laws of Motion, this net force would be:

[tex]F = m\, a = m\, (-A\, \omega^{2})\, \sin(\omega\, t)[/tex].

Divide the restoring force (which is equal to the net force) by displacement to find an expression for the spring constant:

[tex]\begin{aligned}k &= -\frac{F}{x} \\ &= -\frac{m\, (-A\, \omega^{2})\, \sin(\omega\, t)}{A\, \sin(\omega\, t)} \\ &= m\, \omega^{2}\end{aligned}[/tex].

Since [tex]\omega = 2\, \pi\, f[/tex], this equation becomes:

[tex]k = m\, \omega^{2} = (2\, \pi\, f)^{2}\, m[/tex].

Rearrange this equation to find mass [tex]m[/tex]:

[tex]\begin{aligned}m &= \frac{k}{(2\, \pi\, f)^{2}} \\ &= \frac{(242\; {\rm N\cdot m^{-1}})}{(2\, \pi\, (9.7\; {\rm s^{-1}}))^{2}} \\ &\approx 0.0651\; {\rm N\cdot m^{-1}\cdot s^{2}}\end{aligned}[/tex].

Note that [tex]1\; {\rm N} = 1\; {\rm kg\cdot m\cdot s^{-2}}[/tex]. Hence:

[tex]\begin{aligned}m &\approx 0.0651\; {\rm N\cdot m^{-1}\cdot s^{2}} \\ &= 0.065\; ({\rm kg\cdot m\cdot s^{-2}})\, ({\rm m^{-1}})\, ({\rm s^{2}}) \\ &= 0.0651\; {\rm kg\end{aligned}[/tex].

Apply unit conversion and round to the nearest gram:

[tex]m\approx 0.0651\; {\rm kg} \approx 65\; {\rm g}[/tex].