the earth's magnetic field, like any magnetic field, stores energy. the maximum strength of the earth's field is about 7.0 10-5 t. find the maximum magnetic energy stored in the space above a city if the space occupies an area of 3.40 108 m2 and has a height of 1300 m.

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Answer 1

The maximum magnetic energy stored in the space above a city, with an area of 3.40 108 m2 and a height of 1300 m, can be calculated using the equation E = B²V/2, where B is the strength of the Earth's magnetic field (7.0 10-5 t), V is the volume of the space (4.42 10¹⁰ m³), and E is the energy stored. Plugging in the values for B and V, we find the maximum magnetic energy stored in the space above the city to be 6.17 10¹⁵ J.

The Earth's magnetic field is important for providing a protective barrier against dangerous cosmic rays, and for allowing creatures that use magnetoception, such as birds, to navigate.

The Earth's magnetic field is always changing and is generated by electrical currents in the core of the Earth, which are influenced by convection currents and the rotation of the Earth. Even though the energy stored in the Earth's magnetic field is quite small, it still plays an important role in the way the Earth works.

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for an incandescent bulb, initial cost may be high but the energy costs will be low over its life time. (1 point) group of answer choices true false

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True. An incandescent bulb may have a higher initial cost than other types of lightbulbs, but it uses less energy over its lifetime and thus reduces energy costs.


For an incandescent bulb, the given statement is true. In candescent bulbs are traditional bulbs, which use a filament to create light. These bulbs are less efficient, as they waste most of the electricity they use as heat rather than light. As a result, the bulbs are less cost-effective in the long run.

They use up more energy than modern alternatives such as CFLs (compact fluorescent lights) or LEDs (light-emitting diodes). Despite their low initial cost, incandescent bulbs are not recommended for long-term use. They consume more electricity and thus have a greater impact on the environment. Therefore, it is not true that the energy costs of an incandescent bulb will be low over its life time.

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the intensity of sound in a typical classroom is approxiamtely 10^-7 w/m2. what is the sound level for this noise/

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The sound level for this noise is approximately 50 decibels.

Sound level is a logarithmic measure of the ratio between the sound pressure level of a particular sound wave and a reference level. The reference level is typically set at the threshold of human hearing, which corresponds to an intensity of 10^-12 W/m^2. The sound level (measured in decibels, dB) of a sound wave is given by,

L = 10 log10(I/I0)

where I is the intensity of the sound wave and I0 is the reference intensity, which is typically set at 10^-12 W/m^2.

So, for an intensity of 10^-7 W/m^2 in a typical classroom, we can calculate the sound level as,

L = 10 log10(I/I0) = 10 log10(10^-7/10^-12) = 10 log10(10^5) = 50 dB

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determine the intensity of electromagnetic waves from the sun just outside the atmospheres of the earth.

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The intensity of the electromagnetic radiation from the Sun just outside the atmosphere of the Earth is 1.55 x 10-9 W/m2.

The intensity of electromagnetic waves from the sun just outside the atmosphere of the Earth can be calculated using the inverse-square law.

This law states that the intensity of the radiation decreases with the square of the distance from the source. Thus, the intensity of the radiation at the edge of the atmosphere will be lower than that at the surface of the Sun.

The intensity of the radiation, we need to know the distance from the Sun to the Earth. This distance is approximately 93 million miles (150 million kilometers).

The intensity of the radiation at the edge of the atmosphere by taking the inverse-square of this distance, which is approximately 1.55 x 10-9 W/m2.

This is the intensity of the electromagnetic radiation from the Sun just outside the atmosphere of the Earth.

The intensity of the electromagnetic radiation from the Sun just outside the atmosphere of the Earth is 1.55 x 10-9 W/m2.

This is due to the inverse-square law, which states that the intensity of radiation decreases with the square of the distance from the source.

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what is the longest possible wavelength for the traveling waves that can interfere to form a standing wave on this string?

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The longest possible wavelength for a standing wave on a string is: the length of the string itself

This is because, in order to create a standing wave, two traveling waves must interfere with one another and create a wave pattern that is fixed in space.

As the wavelength of the traveling waves increases, the nodes (points of zero displacements) of the standing wave become closer together. Therefore, if the wavelength is equal to the length of the string, the nodes of the standing wave are located at the two ends of the string and the wave pattern remains stationary.

This means that any longer wavelength traveling wave would not be able to interfere and form a standing wave on the string.

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if a 3 solar mass star and a 10 solar mass star formed together in a binary system, which star would evolve off the main sequence first?

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In a binary system with a 3 solar mass star and a 10 solar mass star, the 10 solar mass star would evolve off the main sequence first. This is because more massive stars have shorter lifetimes due to their higher rate of nuclear fusion.

As per the masses of the 3 solar mass star and the 10 solar mass star, the 3 solar mass star would evolve off the main sequence first if they formed together in a binary system. The main sequence is a continuous and distinctive band that appears on plots of stellar color versus brightness. Most stars are found in this band, including the Sun.

The main sequence is the band that represents the stars in the core hydrogen-burning phase. In contrast to the core helium-burning red clump giants and the helium-fusing horizontal branch stars, stars on the main sequence are in a stable state of nuclear fusion.

Because of the higher temperatures inside, more massive stars have a greater rate of nuclear reactions and consume their fuel more quickly. As a result, if a 3 solar mass star and a 10 solar mass star formed together in a binary system, the 3 solar mass star would evolve off the main sequence first.

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what is the torque produced by a force of magnitude 90 n that is exerted perpendicular to and at the end of a 0.5m long wrench

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Torque is a measure of the twisting force that is produced when a force is applied to an object and is defined as the product of the force.

The distance from the pivot point to the point of application of the force, multiplied by the sine of the angle between the force vector and the vector from the pivot point to the point of application of the force.

In this case, the force of 90 N is applied perpendicular to the end of a wrench that is 0.5 m long. Assuming the force is applied at the end of the wrench, the distance from the pivot point to the point of application of the force is 0.5 m. Since the force is perpendicular to the wrench.

The angle between the force vector and the vector from the pivot point to the point of application of the force is 90 degrees. Using the formula for torque, the torque produced by the force is: Torque = force x distance x sin(angle)

Torque = 90 N x 0.5 m x sin(90)Torque = 45 Nm

Therefore, the torque produced by the force of magnitude 90 N that is exerted perpendicular to and at the end of a 0.5m long wrench is 45 Nm.

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calculate the kinetic energy of a ball of mass 50g travelling at 30m/s. how much work will need to bee done to stop the ball?​

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Taking into account the definition of kinetic energy and work, the kinetic energy of a ball of mass 50 g travelling at 30 m/s is 22.5 J and  a work of 22.5 J must be done in an opposite direction to stop the ball.

Definition of kinetic energy

Kinetic energy is defined as the energy associated with bodies that are in motion.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a certain mass and in a position of rest, until it reaches a certain speed. This will remain the same unless there is a change in speed or the body returns to its state of rest by applying a force.

Kinetic energy is represented by the following expression:

Ec = 1/2×m×v²

Where:

Ec is kinetic energy, which is measured in Joules (J).m is mass measured in kilograms (kg).v is velocity measured in meters over seconds (m/s).Definition of work

The kinetic energy theorem states that the work done by the applied net force (sum of all forces) is equal to the change in kinetic energy:

Work= Ec final - Ec original

Kinetic energy of the ball

In this case, you know:

m= 50 g= 0.05 kg (being 1000 g= 1 kg)v= 30 m/s

Replacing in the definition of kinetic energy:

Ec = 1/2× 0.05 g× (30 m/s)²

Solving:

Ec= 22.5 J

The kinetic energy is 22.5 J.

Work to stop the ball

Stoping the ball means bringing the velocity to zero. This is:

Ec final= 1/2× 0.05 g× (0 m/s)²

Ec final= 0

Then, work can be calculated as:

Work= Ec final - Ec initial

Work= 0 - 22.5 J

Work= -22.5 J

This means that a work of 22.5 J must be done in an opposite direction.

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a boat moves at 10.8 m/s relative to the water. if the boat is in a river where the current is 2.00 m/s, how long does it take the boat to make a complete round trip of 1 100 m upstream followed by a 1 100-m trip downstream?

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Time taken for the boat to make a complete round trip of 1 100 m upstream followed by a 1 100-m trip downstream is 200 seconds.

The boat moves at 10.8 m/s relative to the water, and the current is 2.00 m/s. To make a complete round trip of 1 100 m upstream followed by a 1 100-m trip downstream, it would take:

When the boat is moving upstream, it is going against the direction of the current.

Upstream: 1 100 m/ (10.8 m/s - 2.00 m/s) = 102.78 s

When the boat is moving downstream, it is going in the same direction as the current,

Downstream: 1 100 m/ (10.8 m/s + 2.00 m/s) = 97.22 s

Total time taken in going upstream and downstream is the sum of the time calculated in both cases

102.78 s + 97.22 s = 200 s

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based on computer models, when is planetary migration most likely to occur in a planetary system? based on computer models, when is planetary migration most likely to occur in a planetary system? shortly after a stellar wind clears the gaseous disk away late in its history, when asteroids and comets occasionally collide with planets early in its history, when there is still a gaseous disk around the star

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According to computer models, planetary migration is most likely to occur in a planetary system early in its history, when there is still a gaseous disk around the star.

What is planetary migration?

Planetary migration is the process by which a planet changes its orbital position over time. The process is often caused by gravitational interactions with other planets or a planetesimal disk, which causes the planet to migrate inward or outward from its original orbit.

Other factors that can contribute to planetary migration include the late stages of a star's evolution when a stellar wind clears the gaseous disk away and asteroids and comets occasionally collide with planets.

However, early in a planetary system's history, when there is still a gaseous disk around the star, is the most likely time for planetary migration to occur.

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what is the voltage across the 5 ohm resistor when the switch has been in position a for a long time?

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The voltage is V = I × R = 5 ohms.

The voltage across a 5 ohm resistor when the switch has been in position a for a long time is determined by Ohm’s Law.

This law states that the voltage (V) across a resistor is equal to the current (I) through it multiplied by the resistance (R). Therefore, the voltage across the 5 ohm resistor is V = I × R = 5 ohms.

This voltage can also be found by considering the flow of electrons. In a circuit with a battery and a switch, electrons flow from the positive terminal of the battery to the negative terminal.

When the switch is in position a, the 5 ohm resistor is in the path of the electrons and acts as a barrier.

This resistance causes the electrons to slow down and the voltage across the resistor is determined by the amount of this resistance.

The voltage across the 5 ohm resistor when the switch has been in position a for a long time is determined by Ohm’s Law and the amount of resistance the resistor provides. The voltage is V = I × R = 5 ohms.

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alston realizes that he wasn't paying attention when thana ran the simulation for which the charge was set to zero, and asks thana to describe the particle's motion. which response is correct?

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Thana explains that when the charge is zero, the motion of the particle in the simulation is a straight line with a constant velocity.

The direction of the velocity depends on the initial conditions and the force acting on the particle. If there are no other forces acting on the particle, it will continue to move in a straight line with a constant velocity until it encounters another force or object. This is an example of Newton's First Law of Motion, which states that an object at rest will stay at rest and an object in motion will stay in motion with a constant velocity unless acted upon by an external force. If there is a force acting on the particle, it will change direction or speed up or slow down. This is an example of Newton's Second Law of Motion, which states that the force acting on an object is equal to its mass times its acceleration. The direction of the force is in the same direction as the acceleration. When the charge is zero, the particle does not experience any force, so it moves in a straight line with a constant velocity. This is a simple example of how particles can be modeled using physics simulations.

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how much energy is stored by the electric field between two square plates, 9.3 cm on a side, separated by a 2.5- mm air gap? the charges on the plates are equal and opposite and of magnitude 13 nc .

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The electric field stored between two square plates of 9.3 cm on a side and separated by a 2.5 mm air gap is 1110 N/C. This can be calculated using Coulomb's law and the given information.


Coulomb's law states that the electric field is equal to the charge (Q) divided by the permittivity of free space (ε₀) multiplied by the distance (d) squared:

E=Q/(ε₀*d²).
Plugging in the given information,

E=(13 nC)/(8.85 x 10⁻¹² * 0.0025²) = 1110 N/C.
This answer uses Coulomb's law to calculate the electric field stored between two square plates, given the plates' side lengths, air gap width, and charge magnitude.

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find the equivalent capacitance of a 4.20-mf capacitor and an 8.50-mf capacitor when they are connected (a) in series and (b) in parallel

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(a) The equivalent capacitance of the 4.20 µF and 8.50 µF capacitors when connected in series is approximately 4.2017 µF.

(b) The equivalent capacitance of the 4.20 µF and 8.50 µF capacitors when connected in parallel is 12.70 µF.

When two capacitors are connected in series, the equivalent capacitance is given by the formula,

1/Ceq = 1/C1 + 1/C2

where C1 and C2 are the capacitances of the two capacitors.

Substituting the given values,

1/Ceq = 1/4.20 µF + 1/8.50 µF

1/Ceq = 0.238 µF^-1

Ceq = 1 / (0.238 µF^-1)

Ceq = 4.2017 µF (rounded to four significant figures)

When two capacitors are connected in parallel, the equivalent capacitance is given by the formula,

Ceq = C1 + C2

where C1 and C2 are the capacitances of the two capacitors.

Substituting the given values,

Ceq = 4.20 µF + 8.50 µF

Ceq = 12.70 µF

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How is the blue color of a reflection nebula related to the blue color of the daytime sky?
Reflection nebulae look blue for the same reason the sky looks blue. Short wavelengths scatter more easily than long wavelengths.

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The blue color of a reflection nebula is related to the blue color of the daytime sky because both phenomena are caused by the scattering of light.

In the case of the daytime sky, the blue color is due to the scattering of sunlight by the Earth's atmosphere, which causes blue light to be scattered more than other colors, making it the dominant color in the sky. In a reflection nebula, the blue color is also caused by the scattering of light, but this time it is by dust grains in the nebula reflecting light from nearby stars.

The dust grains scatter blue light more effectively than other colors, which gives the nebula its characteristic blue color. Therefore, both the blue color of the sky and the blue color of a reflection nebula are a result of the scattering of light by particles in their respective environments.

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--The complete question is, How is the blue color of a reflection nebula related to the blue color of the daytime sky?--

a track star runs a 400-m race on a 400-m circular track in 45 s. what is his angular velocity assuming a constant speed?

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The angular velocity assuming a constant speed for a track-star who runs 400 m circular track in 45 s is 0.139 radians/s.

To calculate the angular velocity first the circumference of the track is 400 meters.

This means that the angular displacement of the track star during the race is:

θ = s / r

where θ is the angular displacement,

s is the distance traveled by the track star (which is equal to the circumference of the track), and

r is the radius of the circular track.

2.) Since the radius of the circular track is half of its diameter, we have:

r = 400 m / 2 = 200 m

Plugging this into the equation for angular displacement, we get:

θ = 400 m / 200 m = 2π radians

3.) Next, we can use the formula for angular velocity:

ω = θ / t

where ω is the angular velocity and

t is the time it takes for the track star to complete the race.

4.)Plugging in the values we have:

ω = θ / t

ω = 2π radians / 45 s

Therefore, the angular velocity of the track star is:

ω = 0.139 radians/s (rounded to three significant figures)

Therefore, the track star's angular velocity assuming a constant speed is approximately 0.139 radians/s

The angular displacement of the track star is equal to one complete revolution around the circular track, which is equal to 2π radians.

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As shown in the above diagram, a positive charge, Q1 = 2.6 μC, is located at a point, x1 = -3.0 m, and a positive charge, Q2 = 1.4 μC, is located at a point, x2 = +4.0 m.
a. Find the magnitude and direction of the Electric Field at the origin due to charge Q1.
b. Find the magnitude and direction of the Electric Field at the origin due to charge Q2.
c. Find the magnitude and direction of the net Electric Field at the origin.

Answers

a) $$E_1 = \frac{(9.0 \times 10⁹ N m²/C²)(2.6 \times 10⁻⁶C)}{(3.0 m)²} \approx 7.80 \times 10⁵ N/C$$, direction is to the right ; b) $$E_2 = \frac{(9.0 \times 10⁹ N m²/C²)(1.4 \times 10⁻⁶ C)}{(4.0 m)²} \approx 3.94 \times 10⁵ N/C$$, electric field is directed towards point charge so, direction is to the left  c) $$|\vec{E}| = \√{E_1² + E_2²} \approx 8.86 \times 10⁵ N/C$$ and its direction is up.

What is positive charge?

Charge that exists in a body that has fewer electrons than protons is known as positive electrons.

a. To find the electric field at the origin due to charge Q1, we can use the formula for the electric field due to point charge:

$$E_1 = \frac{k Q_1}{r_1²}$$

k is Coulomb constant (k = 9.0 × 10⁹ N m²/C²), Q1 is the charge, and r1 is the distance from the charge to the point where we want to find the electric field.

Q1 = 2.6 μC and r1 = 3.0 m (since x1 = -3.0 m is the distance from Q1 to the origin).

$$E_1 = \frac{(9.0 \times 10⁹ N m^2/C²)(2.6 \times 10⁻⁶C)}{(3.0 m)²} \approx 7.80 \times 10⁵ N/C$$

The electric field is directed away from point charge, so direction of the electric field at the origin due to Q1 is to the right (positive x direction).

b. Similarly, to find the electric field at the origin due to charge Q2, we use the same formula:

$$E_2 = \frac{k Q_2}{r_2²}$$

where Q2 = 1.4 μC and r2 = 4.0 m (since x2 = 4.0 m is the distance from Q2 to the origin).

$$E_2 = \frac{(9.0 \times 10⁹ N m²/C²)(1.4 \times 10⁻⁶ C)}{(4.0 m)²} \approx 3.94 \times 10⁵ N/C$$

The electric field is directed towards point charge, so direction of the electric field at the origin due to Q2 is to the left.

c. $$\vec{E} = \vec{E_1} + \vec{E_2}$$

$\vec{E_1}$ is the electric field due to Q1 and $\vec{E_2}$ is the electric field due to Q2.

net electric field at the origin is: $$|\vec{E}| = \√{E_1² + E_2²} \approx 8.86 \times 10⁵ N/C$$ and its direction is up.

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a copper wire has a resisrtance of 200 ohms. a second copper wire with twice the cross-sectional area and the same lenghthj would have a resisrtance of

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The resistance of a wire is inversely proportional to the cross-sectional area of the wire. So, the resistance of the second wire is one-half of the resistance of the first wire, which is 100 ohms.

Resistance of a copper wire with a cross-sectional area and length.The resistance of a copper wire with a cross-sectional area and length can be calculated using the formula:R=ρL/A

Substituting the given values of resistance R and cross-sectional area A for the first copper wire, we get:R₁ = ρL/A ... (1)where ρ is the resistivity of copper, L is the length of the wire and A is the cross-sectional area of the wire.The resistance of the second copper wire with twice the cross-sectional area A but the same length L can be calculated using the same formula as:R₂ = ρL/2A = (1/2)(ρL/A) ... (2)

Substituting the value of R₁ from equation (1) in equation (2), we get:R₂ = (1/2)(R₁) = 1/2 x 200 = 100Ω

Therefore, the resistance of the second copper wire would be 100 ohms.

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select all that apply select all the stars that would have the same luminosity. (use the stefan-boltzmann law.) presented are the radii and temperatures of five stars compared to the sun.

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According to the Stefan-Boltzmann law, the luminosity of a star is directly proportional to the fourth power of its temperature and its radius squared.

The formula for luminosity is:L = 4πR²σT⁴where L is the luminosity, R is the radius, T is the temperature, and σ is the Stefan-Boltzmann constant (5.67 × 10⁻⁸ W/m²K⁴).To determine which stars would have the same luminosity as the sun, we need to compare their luminosity values using the given data. The radii and temperatures of five stars compared to the sun are as follows:Star A: R = 2R⊙, T = 6000 KStar B: R = R⊙, T = 3000 KStar C: R = 0.1R⊙, T = 6000 KStar D: R = 10R⊙, T = 3000 KStar E: R = 2R⊙, T = 15000 KSubstituting the values in the formula, we get:L⊙ = 4π(1²)(5.67 × 10⁻⁸)(5778⁴) ≈ 3.828 × 10²⁶ Wm¹²Star A: L = 4π(2²)(5.67 × 10⁻⁸)(6000⁴) ≈ 1.84 × 10³³ Wm¹²Star B: L = 4π(1²)(5.67 × 10⁻⁸)(3000⁴) ≈ 6.86 × 10²⁹ Wm¹²Star C: L = 4π(0.1²)(5.67 × 10⁻⁸)(6000⁴) ≈ 6.95 × 10²³ Wm¹²Star D: L = 4π(10²)(5.67 × 10⁻⁸)(3000⁴) ≈ 5.48 × 10³⁴ Wm¹²Star E: L = 4π(2²)(5.67 × 10⁻⁸)(15000⁴) ≈ 5.12 × 10³³ Wm¹²

The luminosity values of the stars are as follows:Star A: L ≈ 1.84 × 10³³ Wm¹²Star B: L ≈ 6.86 × 10²⁹ Wm¹²Star C: L ≈ 6.95 × 10²³ Wm¹²Star D: L ≈ 5.48 × 10³⁴ Wm¹²Star E: L ≈ 5.12 × 10³³ Wm¹²Comparing the luminosity values with that of the sun, we can see that stars A and E would have the same luminosity as the sun.

Therefore, the correct answer is: Stars A and E

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When two unknown resistors are connected in series with a battery, the battery delivers total power Ps and carries a total current of I. For the same total current, a total power Pp is delivered when the resistors are connected in parallel. Determine the value of each resistor. (Use any variable or symbol stated above as necessary.)

Answers

The resistence of each resistor can be calculated by using the equation for resistors in series: R = Ps/I and the equation for resistors in parallel: R = Pp/I.

By substituting the given values for Ps, I and Pp into the equations, we get R1 = Ps/I and R2 = Pp/I. Thus, the value of each resistor can be determined by dividing the total power by the total current.

These equations are based on Ohm's law, which states that the voltage across a resistor is equal to the current through the resistor multiplied by the resistance. By connecting resistors in series or parallel, the overall resistance of the network can be calculated. Knowing the total power and total current, the individual resistances of each resistor can be determined.

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mars has a mass of about 0.1075 times the mass of earth and a diameter of about 0.533 times the diameter of earth. the acceleration of a body falling near the surface of mars is about:group of answer choices0.30 m/s21.4 m/s22.0 m/s23.7 m/s226 m/s2

Answers

Mars has a mass of about 0.1075 times the mass of earth and a diameter of about 0.533 times the diameter of earth.The acceleration of a body falling near the surface of Mars is about 3.7 m/s².

Given, Mars has a mass of about 0.1075 times the mass of Earth and a diameter of about 0.533 times the diameter of Earth.

To find the acceleration of a body falling near the surface of Mars, we can use the formula:

g = GM/r²

where: g = acceleration due to gravity on Mars

M = mass of Mars

r = radius of Mars

We know that mass is proportional to the cube of the radius. So we can say:

Mars/Mass of Earth = (radius of Mars / radius of Earth)³0.1075M/ME

                                = (0.533RE/RE)³0.1075

                                = 0.01514ME

Simplifying this equation, we can say:

M = 0.1075 × MEr = 0.533 × RE

Now, let's calculate the acceleration due to gravity on Mars:

g = GM/r²g

  = (6.674 × 10⁻¹¹) × (0.1075 × ME) / (0.533 × RE)²g

  = 3.7 m/s².

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two balls with masses of 2 kg and 6,3 kg travel toward each other at speeds of f14 and 3 respectively if the balls have a head on inelastic collision and the 2 kg ball recoils with a speed of 3.2 how much kinetic energy

Answers

If the balls have a head-on, inelastic collision and the 2.0-kg ball recoils with a speed of 3.2 m/s, the kinetic energy lost in the collision is 364.6 J.

Using conservation of momentum, we can find the final velocity:

m1 * v1 + m2 * v2 = (m1 + m2) * vf

Solving for vf, we get:

vf = (m1 * v1 + m2 * v2) / (m1 + m2)

= (2.0 kg * 14 m/s + 6.3 kg * 4.0 m/s) / (2.0 kg + 6.3 kg)

= 6.0 m/s

The final total kinetic energy of the system is:

KEf = (1/2) * (m1 + m2) * vf^2

= (1/2) * 8.3 kg * (6.0 m/s)^2

= 112.2 J

The kinetic energy lost in the collision is the difference between the initial and final kinetic energies:

KE lost = KEi - KEf

= 476.8 J - 112.2 J

= 364.6 J

An inelastic collision is a type of collision between two or more objects in which the total kinetic energy of the system is not conserved. In an inelastic collision, some or all of the kinetic energy of the colliding objects is converted into other forms of energy such as heat, sound, or deformation of the objects.

In an inelastic collision, the colliding objects stick together after the collision and move with a common velocity. This is in contrast to an elastic collision, in which the colliding objects bounce off each other and the total kinetic energy of the system is conserved. Inelastic collisions can occur in many different situations, such as in car crashes, when two objects collide and stick together, or when a ball hits a wall and loses some of its kinetic energy due to deformation.

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Two balls with masses of 2.0 kg and 6.3 kg travel toward each other at speeds of 14 m/s and 4.0 m/s, respectively. If the balls have a head-on, inelastic collision and the 2.0-kg ball recoils with a speed of 3.2 m/s, how much kinetic energy is lost in the collision?

two pulse waves of equal and opposite amplitude move toward each other on a cord. after they interfere with each other, what happens to the waves?

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The waves will cancel each other out and no waves will remain. If two waves of the same frequency, but different amplitudes, interfere with each other, the resulting wave will have an amplitude equal to the sum of the two wave amplitudes.

What are pulse waves?

Pulse waves are pressure waves that are created as the heart pumps blood throughout the body. They are detected through pulse points, such as on the wrists, neck, or temples. Pulse waves can be measured using a device called a pulse oximeter, which uses a sensor to detect the pressure of the pulse wave.

Pulse waves can provide information about a person’s heart rate and oxygen saturation levels.

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a 200 ohm, 250 ohm and 1000 ohm resistor are connected in parallel across a source. the source current is 6a. how much is the current that flows through the 200 ohm resisto

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The current that flows through the 200 Ω resistor is 1.56 A.

Given resistance values of 200 Ω, 250 Ω, and 1000 Ω are connected in parallel across a source. The source current is 6 A. We are required to find the current that flows through the 200 Ω resistor.

Recall that when resistors are connected in parallel, the current is divided among them. And the voltage across each resistor is the same. The equivalent resistance of three parallel resistors is given by;

1/Rp = 1/R1 + 1/R2 + 1/R3Rp = (R1 x R2 x R3)/(R1R2 + R1R3 + R2R3)

Put the values into the formula;

Rp = (200 x 250 x 1000)/(200×250 + 200×1000 + 250×1000)

Rp = 52.17 Ω

The total current in the circuit, It = 6 A

From Ohm's Law;

V = IR,

where V is the voltage across each resistor

V1 = V2 = V3V = I×R

Therefore; V = I×Rp

The current flowing through the 200 Ω resistor, I1 = V1/200 = I × Rp/200The current flowing through the 200 Ω resistor, I1 = (6×52.17)/200I1 = 1.56 A

Thus, the current that flows through the 200 Ω resistor is 1.56 A.

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a series circuit is a current divider and a parallel circuit is a voltage divider circuit. select one: a. true b. false

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The given statement " A series circuit is a current divider and a parallel circuit is a voltage divider circuit " is True

In a series circuit, the electric current is the same through each component, and the total current is equal to the sum of the currents through each component. Therefore, the current is divided among the components.

In a parallel circuit, the potential voltage across each component is the same, and the total voltage is equal to the sum of the voltages across each component. Therefore, the voltage is divided among the components.

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if a star is 11 pc away from us, will its apparent visual magnitude be higher or lower than its absolute visual magnitude? what if the star is 5 pc away?

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If a star is 11 pc away from us, its apparent visual magnitude will be lower than its absolute visual magnitude. The star's apparent magnitude would be only 0.38 magnitudes lower than its absolute magnitude.

This is because the apparent magnitude of a star is affected by its distance from us. As the distance increases, the star appears dimmer, and its apparent magnitude decreases.

The distance modulus formula gives us a way to calculate the difference between the apparent and absolute magnitudes of a star:

Distance modulus = 5 * log(distance in parsecs) - 5

For a star that is 11 pc away, the distance modulus is,

Distance modulus = 5 * log(11) - 5 = 1.38

This means that the star's apparent magnitude will be 1.38 magnitudes lower than its absolute magnitude.

If the same star were only 5 pc away from us, the distance modulus would be,

Distance modulus = 5 * log(5) - 5 = 0.38

In this case, the star's apparent magnitude would be only 0.38 magnitudes lower than its absolute magnitude. This means that the star would appear brighter and have a higher apparent magnitude when it is closer to us.

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jeff 60 kg and julia 45 kg are in two separate bumper cars 130 each. jeff was moving at 4 m/s north while julie was going 6 m/s west. julia bounces off going 2 m/s at an angle of 15 s of w. what is the final velocity and direction of jeff car

Answers

Final velocity of Jeff's car is 7.133 m/s south. The direction is 59.3° south of east.

In this issue, we can utilize preservation of energy to track down the last speed and course of Jeff's crash mobile after the impact with Julia's. Before the impact, the energy in the x-heading is zero, and in the y-course, it is 60 kg × 4 m/s = 240 kg⋅m/s north. Julia's force is 45 kg × 6 m/s = 270 kg⋅m/s west.After the crash, the energy in the x-course is rationed. The absolute energy in the x-course is as yet zero, as Julia's force that way is likewise zero. In the y-heading, the absolute force after the crash is 60 kg × vj + 45 kg × 2 m/s sin 15°, where vj is Jeff's last speed in the y-course.Utilizing protection of energy, we can compare the force when the crash in the y-heading:

60 kg × 4 m/s + 45 kg × 6 m/s = 60 kg × vj + 45 kg × 2 m/s sin 15°

Working on this situation, we get:

240 kg⋅m/s + 270 kg⋅m/s = 60 kg × vj + 12.19 kg⋅m/s

Addressing for vj, we get:

vj = (240 kg⋅m/s + 270 kg⋅m/s - 12.19 kg⋅m/s)/60 kg

vj = 7.133 m/s south

Consequently, Jeff's last speed is 7.133 m/s south. To find the course, we can utilize geometry. The point of Jeff's last speed concerning the x-pivot is given by:

θ = tan^-1(vj/4 m/s)

θ = 59.3° south of east

Accordingly, the last speed and heading of Jeff's amusement cart are 7.133 m/s at a point of 59.3° south of east.

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determine the total power delivered to the circuit (i.e., the total power dissipated in the resistors)

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To determine the total power delivered to the circuit (i.e., the total power dissipated in the resistors), you can use the formula:

P = I²R ; where P is the power in watts, I is the current in amperes, and R is the resistance in ohms.

To find the current, you can use Ohm's law:

V = IR

where V is the voltage in volts, I is the current in amperes, and R is the resistance in ohms.

Here's an example:

Suppose you have a circuit with two resistors, R1 and R2, connected in series.

The voltage across the circuit is 10 volts, and the resistances of the two resistors are 2 ohms and 4 ohms, respectively. You can find the total resistance of the circuit by adding the resistances of the two resistors:

R = R1 + R2 = 2 + 4 = 6 ohms

To find the current in the circuit, you can use Ohm's law:

I = V/R = 10/6 = 1.67 amps

Then, you can find the power dissipated in each resistor:

P1 = I²R1 = (1.67)²(2) = 5.56 wattsP2 = I²R2 = (1.67)²(4) = 11.11 watts

And finally, you can find the total power dissipated in the circuit by adding the power dissipated in each resistor:Ptotal = P1 + P2 = 5.56 + 11.11 = 16.67 watts

So the total power delivered to the circuit is 16.67 watts.



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on june 9, 1988, sergei bubka broke the world pole-vaulting record for the 8th time in four years by attaining a height of 6.10 m. how long did it take bubka to return to the ground from the highest part of his vault?

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On june 9, 1988, Sergei Bubka broke the world pole-vaulting record for the 8th time in four years by attaining a height of 6.10 m. It took Bubka 1.11 seconds to return to the ground from the highest part of his vault.

Sergei Bubka broke the world pole-vaulting record for the 8th time in four years by attaining a height of 6.10 m on June 9, 1988. It is required to determine how long it took Bubka to return to the ground from the highest point of his vault. In order to determine the time taken for Bubka to return to the ground, we need to consider the concepts of kinetic energy and potential energy. The pole vaulter gains potential energy during the ascent phase of the vault as he gains altitude. When he reaches the highest point, he has the maximum potential energy. During the descent phase of the vault, the potential energy is converted into kinetic energy.

Based on this principle, we can use the conservation of energy equation to find the time taken by Bubka to return to the ground. The equation for conservation of energy is given as: Potential energy (P.E) = Kinetic energy (K.E)

P.E = mgh where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above the ground.

K.E = 1/2 mv² where v is the velocity of the object.

The velocity of Bubka when he reached the highest point can be assumed to be zero since he had to come to a stop before starting his descent. Therefore, the initial kinetic energy is zero.

P.E at the highest point = K.E at the lowest point

Let t be the time taken by Bubka to return to the ground. We can assume that Bubka moves with uniform acceleration. Using the kinematic equation, we have: v = u + at where u is the initial velocity and a is the acceleration.

When Bubka reaches the ground, his final velocity is zero.

Therefore, we have: v = 0u = at

Substituting the value of u in the equation for K.E, we have: K.E = 1/2 mv² = 1/2 ma²t²

Substituting the value of P.E and K.E in the equation for conservation of energy, we have:

mgh = 1/2 ma²t²

Simplifying, we get: t = sqrt(2h/g)

Substituting the values of h and g, we have:

t = sqrt(2 x 6.10 / 9.81)t = sqrt(1.240)t = 1.11 seconds

Therefore, it took Bubka 1.11 seconds to return to the ground from the highest part of his vault.

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the force of gravity on the side of the earth facing the moon is the force of gravity acting on the center of the earth group of answer choices greater than smaller than equal to

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The force of gravity on the side of the earth facing the moon is greater than the force of gravity acting on the center of the earth.

This is because of the gravitational attraction between the earth and the moon.

The moon’s gravity pulls on the side of the earth that is closer to it, resulting in a larger gravitational force on that side than on the center of the earth. The size of the force on the side of the earth is slightly more than double that at the center, due to the inverse square law.

Thus, the force of gravity at the side of the earth facing the moon is greater than the force of gravity acting on the center of the earth.

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The force of gravity on the side of the earth facing the moon is the force of gravity acting on the center of the earth

greater than

smaller than

equal to

what is the mass, in units of me (the mass of the earth), of a planet with twice the radius of earth for which the escape speed is twice that for earth?

Answers

The mass, in units of me (the mass of the earth), of a planet with twice the radius of the earth for which the escape speed is twice that of the earth is 8 me.

The amount of matter in an object is referred to as mass. Mass is expressed in terms of the unit kilogram in the International System of Units (SI).

The escape velocity is defined as the minimum velocity required for an object to leave the gravitational influence of another object. For example, if a ball is thrown from the surface of the earth at a speed of 11.2 km/s (40,320 km/h), it will escape the earth's gravitational pull and continue into space.

The formula for escape velocity is given by:

  v=√(2GM/r)

Where, v is the escape velocity, G is the gravitational constant, M is the mass of the planet, and r is the radius of the planet.

The formula for mass:

  m = v²r/Gm = (2v)²(2r)/GMm = 8r/G

Therefore, the mass, in units of me (the mass of the earth), of a planet with twice the radius of earth for which the escape speed is twice that of the earth is 8 me.

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