Compare the relationship between weathering and erosion to the relationship between convergent and divergent boundaries.

To calculate the angular momentum of the Earth in its orbit around the Sun, we use the formula:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

The moment of inertia of the Earth in its orbit around the Sun can be approximated as a point mass at the center of the orbit, so we have:

I = mr^2

where m is the mass of the Earth, and r is the radius of its orbit around the Sun.

The angular velocity of the Earth in its orbit around the Sun can be calculated as:

ω = v/r

where v is the velocity of the Earth in its orbit around the Sun.

Using the values of the mass of the Earth (m = 5.97 × 10^24 kg), the radius of its orbit around the Sun (r = 1.50 × 10^11 m), and the velocity of the Earth in its orbit around the Sun (v = 2.98 × 10^4 m/s), we have:

I = (5.97 × 10^24 kg) (1.50 × 10^11 m)^2 = 1.08 × 10^40 kg m^2

ω = (2.98 × 10^4 m/s) / (1.50 × 10^11 m) = 1.99 × 10^-7 rad/s

Therefore, the angular momentum of the Earth in its orbit around the Sun is:

L = Iω = (1.08 × 10^40 kg m^2) (1.99 × 10^-7 rad/s) = 2.15 × 10^33 kg m^2/s

To calculate the angular momentum of the Earth spinning on its own axis, we use the same formula:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

The moment of inertia of the Earth spinning on its own axis can be approximated as a solid sphere, so we have:

I = (2/5)mr^2

where m is the mass of the Earth, and r is the radius of the Earth.

The angular velocity of the Earth spinning on its own axis is:

ω = 2π/T

where T is the period of rotation of the Earth.

Using the values of the mass of the Earth (m = 5.97 × 10^24 kg), the radius of the Earth (r = 6.37 × 10^6 m), and the period of rotation of the Earth (T = 24 hours = 8.64 × 10^4 s), we have:

I = (2/5) (5.97 × 10^24 kg) (6.37 × 10^6 m)^2 = 8.03 × 10^37 kg m^2

ω = 2π / (8.64 × 10^4 s) = 7.27 × 10^-5 rad/s

Therefore, the angular momentum of the Earth spinning on its own axis is:

L = Iω = (8.03 × 10^37 kg m^2) (7.27 × 10^-5 rad/s) = 5.84 × 10^33 kg m^2/s

The angular momentum of the Earth in its orbit around the Sun is approximately 2.76 × 10^40 kg m^2/s, and the angular momentum of the Earth spinning on its axis is approximately 7.06 × 10^33 kg m^2/s.

To find out how many times larger the angular momentum of the Earth in its orbit is compared to the angular momentum of the Earth spinning on its axis, we can simply divide the value obtained in part (a) by the value obtained in part (b):

2.76 × 10^40 kg m^2/s ÷ 7.06 × 10^33 kg m^2/s ≈ 3.91 × 10^6

Therefore, the angular momentum of the Earth in its orbit around the Sun is approximately 3.91 million times larger than the angular momentum of the Earth spinning on its axis.

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Answer:

Na+ca+CO=NACo^3

Explanation:

NA Burns violently with explosions that may spatter the material. Used for making gasoline additives, electric power cable, sodium lamps, other chemicals.

Answer:

What class is this for, I think I have done this before.

The y-component of the force on the particle at the given position is 120 N.

The electric force on the particle is determined by applying Coulomb's law and work-energy theorem as shown below;

Fd = W

Where;

F = W/d

F = 60/0.5

F = 120 N

Thus, the y-component of the force on the particle at the given position is 120 N.

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Answer:

The formation of reddish brown ferric oxide on iron by low temperature oxidation in the presence of water .

make me as brilliant answer

The house uses a total of 3.75 kW of power, with 1.5 kW of that power being supplied by the solar panels.

Solar power is generated by converting the energy from sunlight into electricity using solar panels. Solar panels are made up of photovoltaic cells, which are made of materials that can absorb sunlight and convert it into electricity.

When sunlight hits a solar panel, it causes the electrons in the photovoltaic cells to become excited and move around, generating an electric current.

To find the total amount of power used by the house, you can use the following formula:

Power used by house = Solar power generated / Percentage of power supplied by solar panels

Plugging in the given values, you get:

Power used by house = 1.5 kW / 0.4

= 3.75 kW

This means that the house uses a total of 3.75 kW of power, with 1.5 kW of that power being supplied by the solar panels.

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Answer:

It's d

Explanation:

Answer: D

Explanation:

Force of gravity

Let's call the length of the race track as "x".

To solve for "x", we can use the formula:

\(\implies\text{ Time} = \dfrac{\text{ Distance}}{\text{ Speed}}\)

Since the race ends in a tie, we know that Jan and Mary both took the same amount of time to run the race. Let's call this time "t".

\(\implies \text{ Time} = \dfrac{\text{ Distance}}{\text{ Speed}}\)

\(\implies t = \dfrac{x - 25}{7.5}\)

\(\implies\text{ Time} = \dfrac{\text{ Distance}}{\text{ Speed}}\)

\(\implies t = \dfrac{x}{8.0}\)

Since both expressions represent the same time, we can set them equal to each other and solve for "x":

\(\implies \dfrac{x - 25}{7.5} = \dfrac{x}{8.0}\)

Multiplying both sides by 60 (the least common multiple of 7.5 and 8.0) to get rid of the decimals:

\(\implies 8(x - 25) = 7.5x\)

\(\implies 8x - 200 = 7.5x\)

\(\implies 0.5x = 200\)

\(\implies x = 400\)

\(\therefore\) The length of the race track is 400 meters.

\(\blue{\overline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}\)

Answer:

The answer is B

Explanation:

the rate at which someone or something moves or operates or is able to move or operate.

Let's explain why Common Emitter (CE) configuration is preferred over Common Collector (CC) for amplification purposes.

Common Emitter amplifier is an amplifier that offers high current gain, medium input resistance and high output resistance.

Common Collector is another type of amplifier where the input signal is applied to the base and the ouput signal is taken from the emitter.

The Common Emitter configuration is preferred over the Common Collector configuration because the Common Emitter offers a high power gain(high voltage and current gain), and has a lower resistance when compared to the Common Collector configuration.

ANSWER:

The common Emitter configuration has higher voltage and current gain when compared to the Common Collector configuration.

The pressure will become 2.68 atm when expressed in atmospheres.

The act of shifting a quantity's measurement between two units of measurement is known as unit conversion. Usually, multiplicative conversion factors are used, which change the value of the measured amount without altering its effects.

The metric system typically requires moving a decimal point in order to translate from one unit to another.

The given data is 39.5 \(lb/in^2\)

we have to convert it into atmospheres

The pressure is :

39.5  \(lb/in^2\) ×  \(\frac{1 atm}{ 14.7 lb/in^2}\)

because the conversion factor is 14.7  \(lb/in^2\)

thus the given pressure becomes ;

Pressure = 2.68 atm

The pressure will become 2.68 atm when expressed in atmospheres.

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Based on my observations in this activity, if the hydrogen peroxide solution were diluted by half, it would likely result in a decrease in the reaction rate of the catalase enzyme.

When the hydrogen peroxide concentration is high, the enzyme catalase works quickly to break down the peroxide into water and oxygen. However, if the concentration is diluted, there are fewer hydrogen peroxide molecules available for the enzyme to react with, which means the reaction rate will be slower.

This is because the rate of a chemical reaction is typically proportional to the concentration of the reactants. When the concentration of the reactants is higher, there are more molecules available to react with each other, which means the reaction will occur more quickly. When the concentration is lower, there are fewer molecules available to react, which means the reaction will occur more slowly.

In this case, if the hydrogen peroxide solution were diluted by half, there would be half as many hydrogen peroxide molecules available for the catalase enzyme to react with, which would result in a slower reaction rate.  

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Part A: At terminal velocity, the net force on the skydiver is zero, so the drag force Fd is equal in magnitude but opposite in direction to the force of gravity Fg. Thus, we have:

Fd = mg

The work done on the skydiver by the drag force over the distance d is:

Wd = Fd d = (mg) d

Substituting the equation for terminal velocity:mg = (1/2)ρAv²Cd

where ρ is the density of air, A is the cross-sectional area of the skydiver, and Cd is the drag coefficient.

Solving for m:

m = (1/2)ρAv²Cd / g

Substituting into the expression for Wd:

Wd = [(1/2)ρAv²Cd / g] d

Part B: At terminal velocity, the power supplied by the drag force is equal in magnitude but opposite in sign to the power lost to air resistance. Since the net power is zero, the power supplied by the drag force is:

Pd = Fd v = mgv

Substituting the expression for m:

Pd = [(1/2)ρAv²Cd / g] g d = (1/2)ρAv³Cd / d

where we have used the equation for terminal velocity to eliminate the variable v.

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(a) The diameter of the lens aperture is approximately 0.0224 mm.

(b) The f-number is approximately 12.05.

(a) The radius of the Airy disk, which represents the diffraction pattern formed by the lens aperture, is given as 0.0112 mm. The diameter of the aperture can be calculated by multiplying the radius by 2:

Diameter = 2 * Radius = 2 * 0.0112 mm = 0.0224 mm.

(b) The f-number is the ratio of the focal length of the lens to the diameter of the lens aperture. Given the focal length of 135 mm and the aperture diameter of 0.0224 mm, we can calculate the f-number:

f-number = Focal Length / Aperture Diameter = 135 mm / 0.0224 mm ≈ 12.05.

Therefore, the diameter of the lens aperture is approximately 0.0224 mm, and the f-number is approximately 12.05. The f-number is commonly used to represent the speed of a lens, with smaller values indicating larger aperture openings and faster lenses.

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FAST

MEDium

slow

.....

The charged balloon sticks to a wall due to electrostatic force between the charged balloon and the wall.

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Answer:

(A). The force acting on the rocket is mg.

(B). The initial upward acceleration of the rocket is 6.2 m/s²

Explanation:

Given that,

Mass of rocket = 5 kg

Force = 80 N

Time = 4 sec

(A). We need to draw the figure

The force acting on the rocket at each of the intervals is shown in figure.

The force acting on the rocket is mg.

(B). We need to calculate the initial upward acceleration of the rocket

Using balance equation

\(ma=F-mg\)

\(a=\dfrac{F-mg}{m}\)

Put the value in the equation

\(a=\dfrac{80-5\times9.8}{5}\)

\(a=6.2\ m/s^2\)

Hence, (A). The force acting on the rocket is mg.

(B). The initial upward acceleration of the rocket is 6.2 m/s²

→ The charge in electron charge Q(e) is equal to the charge in coulombs Q(C) times 6.24150975⋅1018:

1C = 6.24150975⋅10¹⁸e

Q(e) = Q(C) × 6.24150975 × 10¹⁸

Q(e) = 96,500 × 6.24150975 × 10¹⁸

Q(e) = 6.02305690875 × 10²³

which is round off to

→ 6.03 × 10²³ electrons

The acceleration of a body can be calculated using the following equation:

s = ut + ½at²

Where;

400 = 0 × 20 + ½ × a × 20²

400 = 0 + 200a

a = 2m/s²

Force acting on the truck can be calculated by multiplying the mass of the truck by its acceleration as follows:

Force = 7000kg × 2m/s²

Force = 14,000N

Therefore;

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Answer:

I'm not sure what your asking for

reaction and law

Explanation:

the newton third law of motion

Answer:

jgfsdghjjuk

Explanation:

khhnghjj

Answer:

I am 98% sure that the answers are B and A

Explanation:

Answer:

Rayleigh scattering is caused by small molecules in the air, and it scatters light more heavily at a shorter wavelengths.

a) The expression to determine the velocity of cart A is \(v_{A} = 2\cdot \sqrt{g\cdot H}\).

b) The expression to determine the velocity of cart B is

\(v_{B} =\sqrt{2\cdot g\cdot H }\).

c) The final velocity of the carts after the collision is \(v = \left(\frac{1-\sqrt{2}}{3} \right)\cdot \sqrt{g\cdot H}\).

d) The collision is inelastic since a part of the energy of the entire system is lost when they stick together.

In this question we shall apply principle of energy conservation and principle of linear momentum conservation to model an inelastic collision between two carts.

a) The combination of cart and ramp can be considered a conservative system as there are no non-conservative forces (i.e. friction), the final velocity of cart A (\(v_{A}\)) is related to the change in gravitational potential energy:

\(\frac{1}{2}\cdot M\cdot v_{A}^{2} = M\cdot g \cdot 2\cdot H\) (1)

Now we clear \(v_{A}\) and simplify the resulting expression:

\(v_{A} = 2\cdot \sqrt{g\cdot H}\)

The expression to determine the velocity of cart A is \(v_{A} = 2\cdot \sqrt{g\cdot H}\). \(\blacksquare\)

b) We apply the same approach used in part b) to find the final velocity:

\(\frac{1}{2}\cdot (2\cdot M) \cdot v_{B}^{2} = (2\cdot M)\cdot g \cdot H\) (2)

Now we clear \(v_{B}\) and simplify the resulting expression:

\(v_{B} =\sqrt{2\cdot g\cdot H }\)

The expression to determine the velocity of cart B is

\(v_{B} =\sqrt{2\cdot g\cdot H }\). \(\blacksquare\)

c) The final velocity (\(v\)system is determined by principle of linear momentum conservation:

\(3\cdot M\cdot v = M\cdot 2\cdot \sqrt{g\cdot H}-2\cdot M\cdot \sqrt{2\cdot g\cdot H}\)

\(3\cdot v = 2\cdot (1-\sqrt{2})\cdot \sqrt{g\cdot H}\)

\(v = \left(\frac{1-\sqrt{2}}{3} \right)\cdot \sqrt{g\cdot H}\)

The final velocity of the carts after the collision is \(v = \left(\frac{1-\sqrt{2}}{3} \right)\cdot \sqrt{g\cdot H}\). \(\blacksquare\)

d) The collision is inelastic since a part of the energy of the entire system is lost when they stick together. \(\blacksquare\)

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Answer: In places of hot climate it is advised that the outer wall of houses be painted white because white color reflects heat and the houses do not heat up too much.

When a 2.80 cm high insect is positioned 1.30 m away from a lens with a focal length of 135 mm, the resulting image is located 10.07 cm from the lens, has a height of 0.22 cm, and is a virtual and upright image.

(ii) (a) Given:

Object height (h₁) = 2.80 cm

Object distance (u₁) = 1.30 m = 130 cm

Focal length of the lens (f₁) = 135 mm = 13.5 cm

Using the lens formula: 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.

Substituting the given values:

1/13.5 = 1/v - 1/130

Solving for v, the image distance formed by the lens:

1/v = 1/13.5 + 1/130

1/v = (10 + 1)/135

1/v = 11/135

v = 135/11 ≈ 12.27 cm

M₁ = -v/u₁

M₁ = -(12.27/130)

M₁ ≈ -0.094

formula: h₂ = |M₁| * h₁

h₂ = |-0.094| * 2.80

h₂ ≈ 0.263 cm

The height of the image (h₂) is approximately 0.263 cm.

Therefore, the image formed by the lens is real, inverted, and has a height of approximately 0.263 cm. It is located at a distance of approximately 12.27 cm from the lens

(ii) (b) If the focal length of the lens is changed to 135 mm (0.135 m), we can repeat the calculations using the new focal length.

Object height (h₁) = 2.80 cm

Object distance (u₁) = 1.30 m = 130 cm

Focal length of the lens (f₁) = 0.135 m = 13.5 cm

1/13.5 = 1/v - 1/130

Solving for v, the image distance formed by the lens:

1/v = 1/13.5 + 1/130

1/v = (10 + 1)/135

1/v = 11/135

v = 135/11 ≈ 12.27 cm

The image distance (v) and the magnification (M) remain the same as in part (a) because the object distance (u₁) and the focal length (f₁) are unchanged. Therefore, the image would still be located at approximately 12.27 cm from the lens, have a height of approximately 0.263 cm, and be real, inverted.

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