an asteroid exerts a 360-n gravitational force on a nearby spacecraft. if the spacecraft moves to a position three times as far from the center of the asteroid, the force will be

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Answer 1

The force between the asteroid and the spacecraft will be 40 N when the spacecraft moves to a position three times as far from the center of the asteroid.

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

F = G(m₁m₂) / r²

where G is the gravitational constant.

In this problem, the asteroid exerts a gravitational force of 360 N on the spacecraft when they are at a certain distance r from each other. When the spacecraft moves to a position three times as far from the center of the asteroid, its distance from the asteroid will be 3r. To calculate the new force between them, we can use the same formula and plug in the new distance:

F' = G(m1m2) / (3r)^2

F' = G(m1m2) / 9r^2

Since the masses of the asteroid and spacecraft are constant, we can divide the second equation by the first to find the ratio of the new force to the original force:

F' / F = (G(m₁m₂) / r²) / 9r²) / (G(m₁m₂) / r²)

F' / F = (1 / 9)

F' = (1 / 9) * F

F' = (1 / 9) * 360 N

F' = 40 N

Therefore, the force will be 40 N.

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Related Questions

Suppose you were to compare three stars with the same surface temperature. If star A is a giant star, star B is a supergiant star, and star C is a main sequence star, order the three stars in terms of increasing radius. a. Star C, Star A, Star B b. Star B, Star A, Star C c. Star A, Star C, Star B d. Star B, Star C, Star A

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If star A is a giant star, star B is a supergiant star, and star C is a main sequence star, the order of the three stars in terms of increasing radius is Star A, Star C, Star B.

A giant star is a luminous star that is considerably larger and brighter than the sun. The distinction between giant and dwarf stars is primarily determined by their luminosity, and giant stars are more luminous. They are not, however, larger in diameter than dwarf stars. Their size is the outcome of a high luminosity-to-mass ratio.

A supergiant star is a massive star with a luminosity that is many times greater than that of a giant star. As a result, a supergiant star is much larger than a giant star. However, supergiant stars have a similar surface temperature as giant stars.

Sequence stars are stars that spend most of their lives in the primary sequence of stars. A main-sequence star is a star that is in the hydrogen-burning phase of its evolution. It is in a state of hydrostatic equilibrium, meaning that the gravitational force holding the star together is balanced by the pressure generated by the thermonuclear fusion taking place in its core.

The stars will have the following order in terms of increasing radius: Star A, Star C, Star B if star A is a giant star, star B is a supergiant star, and star C is a main sequence star, and they all have the same surface temperature.

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what principle states that the buoyant force experienced by an object is exactly equal to the weight of the fluid displaced?

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The principle that states that the buoyant force experienced by an object is exactly equal to the weight of the fluid displaced is known as Archimedes' Principle. What is Archimedes' Principle? Archimedes' Principle is a scientific law that explains how objects behave in fluids (liquids and gases).

The buoyant force of an object in a fluid is equal to the weight of the fluid displaced by the object according to this principle. This principle is valid for any fluid and any object as long as the buoyancy and weight of the object and fluid are calculated correctly.

The force that causes objects to float or sink in fluids is known as buoyancy. The buoyant force on an object is the net upward force exerted by the fluid in which the object is submerged.

When an object is immersed in a fluid, the fluid exerts an upward force on the object. This buoyant force opposes the weight of the object and causes it to float if the buoyant force is greater than the weight of the object.

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the rotational speed of a flywheel increases by 40%. by what percent does its rotational kinetic energy increase? explain your answer.

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The rotational kinetic energy of a flywheel increases by 80% when its rotational speed increases by 40%. This is because the rotational kinetic energy of a flywheel is directly proportional to the square of its angular velocity.


The rotational speed of a flywheel increases by 40%. The percentage increase in its rotational kinetic energy is approximately 96.8%. Suppose the initial rotational speed of the flywheel is n1 and the initial rotational kinetic energy is K.E.1. After the speed of the flywheel is increased by 40 percent, the new speed is n2 = n1 + 0.4n1 = 1.4n1.

Then the new kinetic energy K.E.2 of the flywheel is given by K.E.2 = (1/2)I(n2^2)where I is the moment of inertia of the flywheel.Since n2 = 1.4n1, we have [tex]K.E.2 = (1/2)I(1.96n1^2) = 0.98I(n1^2).[/tex].

Therefore, the percentage increase in the rotational kinetic energy of the flywheel is approximately 96.8%.

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a load of 12 kg stretches a spring to a total length of 15 cm, and a load of 30 kg stretches it to a length of 18 cm. find the natural (unstretched) length of the spring.

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The natural length of the spring is therefore 12.97 cm.

The natural length of the spring is found by calculating the spring constant using the Hooke's law formula. Spring constant (k) = Force (F) / extension (x). The natural length of the spring refers to the length of the spring when it is not carrying any load. Hooke's law states that the force required to extend or compress a spring by a distance x is proportional to that distance. Mathematically, F=kx, where F is the force applied, x is the displacement from the equilibrium position, and k is the spring constant. To find the natural length of the spring, we need to calculate the spring constant.

To do this, we use the data given in the problem. A load of 12 kg stretches the spring to a total length of 15 cm. We can find the force applied by multiplying the load by the acceleration due to gravity (g), which is 9.8 m/s^2. Thus, F = mg = 12 * 9.8 = 117.6 N. The extension of the spring is given as x = 15 cm - x0, where x0 is the natural length of the spring. Thus, x = 0.15 m - x0. Substituting these values into Hooke's law, we get: k = F/x = 117.6/(0.15 - x0)

Similarly, when a load of 30 kg stretches the spring to a length of 18 cm, we can find the force applied as F = mg = 30 * 9.8 = 294 N. The extension is given as x = 0.18 m - x0. Substituting these values into Hooke's law, we get: k = F/x = 294/(0.18 - x0)

Now we have two equations for k, so we can set them equal to each other: 117.6/(0.15 - x0) = 294/(0.18 - x0) Cross-multiplying and simplifying, we get: 117.6(0.18 - x0) = 294(0.15 - x0) 21.168 - 117.6x0 = 44.1 - 294x0 176.4x0 = 22.932 x0 = 0.1297 m

The natural length of the spring is therefore 12.97 cm.

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a gun fires a bullet vertically into a 1.40-kg block of wood at rest on a thin horizontal sheet.if the bullet has a mass of 26.8 g and a speed of 230 m/s , how high will the block rise into the air after the bullet becomes embedded in it?

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The block will rise to a height of approximately 4.36 cm after the bullet becomes embedded in it.

We can use the principle of conservation of momentum to solve this problem. The total momentum of the system (bullet + block) before the collision is,

p_before = m_bullet * v_bullet

where m_bullet is the mass of the bullet and v_bullet is its speed.

After the collision, the bullet becomes embedded in the block, so the total mass of the system is,

m_total = m_bullet + m_block

The velocity of the combined bullet-block system after the collision can be calculated using the conservation of momentum,

p_before = p_after

m_bullet * v_bullet = (m_bullet + m_block) * v_after

where v_after is the velocity of the combined bullet-block system after the collision.

Solving for v_after,

v_after = (m_bullet * v_bullet) / (m_bullet + m_block)

Now, we can calculate the kinetic energy of the bullet-block system just after the collision,

KE_after = (1/2) * (m_bullet + m_block) * v_after^2

The initial kinetic energy of the bullet is,

KE_before = (1/2) * m_bullet * v_bullet^2

The difference between these two energies represents the energy that has been transferred to the block,

delta_KE = KE_before - KE_after

This energy is used to raise the block to a certain height h. If we assume that all of this energy is converted into potential energy, then we can write,

delta_KE = m_block * g * h

where g is the acceleration due to gravity.

Solving for h,

h = delta_KE / (m_block * g)

Substituting the expressions for delta_KE, m_block, v_bullet, and v_after,

h = [(1/2) * m_bullet * v_bullet^2] / [(m_bullet + m_block) * g]

Substituting the given values,

h = [(1/2) * 0.0268 kg * (230 m/s)^2] / [(0.0268 kg + 1.40 kg) * 9.81 m/s^2] = 0.0436 m

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Two large parallel metal plates carry opposite charges. They are separated by 10 cm and p. D of 500 volts is applied on them. What is the magnitude of electric field strength between them? compute the work done by the field on a change of 2x10^-9 as it moves from higher to lower part?

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(a) The magnitude of electric field in the region between the plates is [tex]\mathbf{9 , 2 5 0}$ $\mathrm{V} / \mathrm{m}$.[/tex]

(b) The magnitude of the force the field exerts on a particle with the given charge i[tex]s $2.22 \times 10^{-5} \mathrm{~N}$.[/tex]

(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is[tex]$8.88 \times 10^{-7} \mathrm{~J}$.[/tex]

(d) the change of the potential energy is[tex]$8.88 \times 10^{-7} \mathrm{~J}$.[/tex]

(a) The magnitude of electric field in the region between the plates is calculated as;

[tex]$$\begin{aligned}& E=\frac{V}{d} \\& E=\frac{370}{40 \times 10^{-3}} \\& E=9,250 \mathrm{~V} / \mathrm{m}\end{aligned}$$[/tex]

(b) The magnitude of the force the field exerts on a particle with the given charge is calculated as follows;

[tex]$$\begin{aligned}& F=E q \\& F=9,250 \times 2.4 \times 10^{-9} \\& F=2.22 \times 10^{-5} \mathrm{~N}\end{aligned}$$[/tex]

(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is calculated as follows;

[tex]$$\begin{aligned}& W=F d \\& W=2.22 \times 10^{-5} \times 40 \times 10^{-3} \\& W=8.88 \times 10^{-7} \mathrm{~J}\end{aligned}$$[/tex]

(d) the change of the potential energy is calculated as;

[tex]$$\begin{aligned}& \Delta U=q \Delta V \\& \Delta U=q\left(V_1-V_2\right)\end{aligned}$$$$\text { DeltaU }=2.4 \times 10^{-9}(370)$$$$\Delta U=8.88 \times 10^{-7} \mathrm{~J}$$[/tex]

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Full Question: Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by a distance of 40.0 mm, and the potential difference between them is 370 V

A. What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?

B. What is the magnitude of the force this field exerts on a particle with a charge of 2.40 nC ?

C. Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower.

D. Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

g a research rocket is launched from boulder straight towards the south. how would the coriolis effect change the path of the rocket?

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For a rocket launched southward from Boulder, the Coriolis effect would cause it to drift to the east, leading to a curved flight path rather than a straight one.

The Coriolis effect is an important force to consider when launching a research rocket from Boulder. The Coriolis effect is the result of Earth's rotation and will cause any object moving along the surface of the Earth to veer to the right in the Northern hemisphere and to the left in the Southern hemisphere.

This effect is most noticeable for objects traveling long distances, such as a rocket. As it continues to fly south, the Coriolis force will continue to act upon it, increasing the curvature of its path. The magnitude of the Coriolis force depends on the speed of the object and its distance from the poles. Therefore, the more time the rocket has to travel, the more it will be deflected from its intended path.

The Coriolis effect is an important factor to consider for any research rocket launch. It has the potential to affect the accuracy and success of the mission and must be taken into account when planning a launch trajectory.

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Complete Question:

A research rocket is launched from Boulder straight towards the south. How would the Coriolis effect change the path of the rocket?

calculate the force required to stop a car of mass 1400 kg in 2 seconds if it is moving with a velocity of 10 m/s.

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The force required to stop a car of mass 1400 kg in 2 seconds if it is moving with a velocity of 10 m/s is 7000 N in the opposite direction to the car's motion.

Calculate the force required to stop a car of mass 1400 kg in 2 seconds if it is moving with a velocity of 10 m/s.

To solve the given problem, we can use the equation:

F = (m * Δv) / Δt

where F = force

required to stop the carm = mass of the car Δv = change in velocity = final velocity - initial velocityΔt = time taken to stop the car.

Given, mass of the car, m = 1400 kg Initial velocity, u = 10 m/s Final velocity, v = 0 m/s Time taken to stop, t = 2 seconds Therefore, Δv = v - u = 0 - 10 = -10 m/s

Substituting the given values in the above equation, we get:

F = (m * Δv) / Δt = (1400 kg * (-10 m/s)) / (2 s) = -7000 N

Here, the negative sign indicates that the force required to stop the car is acting in the opposite direction to the car's motion.

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what would its landing speed have been in the absence of air resistance? express your answer using two significant figures.

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The landing speed of the ball in the absence of air resistance would be 14 m/s.

The landing speed of an object in the absence of air resistance can be calculated by considering the conservation of energy.

The initial energy of the object will be equal to the final energy of the object when it reaches the ground.

A ball falling from a height h with an initial velocity u.

The gravitational potential energy of the ball is given by mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the ball.

The kinetic energy of the ball is given by 1/2 mu², where u is the initial velocity of the ball.

At the ground level, the gravitational potential energy of the ball will be zero, and the kinetic energy of the ball will be given by 1/2 mv², where v is the velocity of the ball when it reaches the ground.



mgh + 1/2 mu² = 1/2 mv²

Solving for v, we get:

v = sqrt(2gh + u²)

In the absence of air resistance, the ball will continue to fall with an acceleration of g. Therefore, we can assume that the initial velocity u is equal to zero. Thus, the equation reduces to:

v = sqrt(2gh)

g = 9.8 m/s², we can calculate the landing speed of the ball for a given height h. For example, if the ball is dropped from a height of 10 meters, then the landing speed of the ball will be:

v = sqrt(2gh) = sqrt(2*9.8*10) = 14 m/s

Therefore, the landing speed of the ball in the absence of air resistance would be 14 m/s.

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What is the transfer of thermal energy called?

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Answer:

Conduction

Explanation:

The process by which heat energy is transmitted through collisions between neighboring atoms

a fragment of a current-carrying wire has a cross-sectional area that increases as shown. 1) if the current that flows through the wire is uniform, where is the drift velocity the greatest?

Answers

According to the given statement, if the current that flows through the wire is uniform, the drift velocity is the greatest at the section of wire with diameter d.

As the current is uniform throughout the wire, so the current through a given cross-sectional area is the same. Also, the current density, J is given by:

J = I/A

where I is the current and A is the cross-sectional area of the wire. Thus, if the area of the cross-section of the wire is more, the current density will be less. The current density is inversely proportional to the area of the wire, i.e. J ∝ 1/A. Hence, the drift velocity is inversely proportional to the current density, i.e. v[tex]_d[/tex] ∝ 1/J.

Thus, the drift velocity is greater where the cross-sectional area is less. So, the drift velocity is greater at the section of wire with diameter d.

So, the answer is at the section of wire with diameter d

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a star simultaneously emits red light, blue light, x-rays, and radio waves in the direction of the earth. which will arrive first?

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The answer is that the radio waves will arrive first at the Earth when a star emits red light, blue light, x-rays, and radio waves.

This is due to the fact that radio waves are long-wavelength electromagnetic radiation. As a result, they are less likely to be impeded or absorbed by the intervening space medium, and they can propagate without being affected by any other disturbances in the cosmos.

Furthermore, radio waves are not influenced by the earth's atmosphere, which is responsible for interfering with the passage of light rays to the surface of the earth. In other words, radio waves can traverse enormous distances in space without being obstructed or attenuated by any physical barrier.


Light rays, on the other hand, propagate via a straight line, which is known as the line of sight. Light rays may be deflected or absorbed by cosmic dust, gas clouds, or other materials found in interstellar space. This may cause them to travel in different directions, which might cause them to be redirected from their initial path. As a result, light rays must contend with these obstacles before reaching the earth, which may cause them to be weakened or distorted by the time they arrive.


Similarly, X-rays are also electromagnetic radiation but they are absorbed by interstellar matter. They are also affected by magnetic fields, and they might be redirected from their path as a result of the interstellar medium. This might cause them to be slowed down and travel a longer distance, making their journey longer.

Thus, radio waves will arrive first because of their long wavelength and low interaction with cosmic matter.

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how do extrusive igneous rocks form

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Answer:

igneous rock is produced when magma exits and cools above (or very near) the Earth's surface. These are the rocks that form at erupting volcanoes and oozing fissures.

two balls are connected to 60-cm-long light strings and the other ends of the strings are fixed together as shown in the figure. one of the balls has a mass of 2.0 kg and is raised up and to the right until it is 12.0 cm higher than the other ball, which has a mass of 3.0 kg. the upper ball is released from rest and sticks to the lower ball when they collide. for the subsequent motion find the:

Answers

According to the question the speed of the balls just before they collide is 1.81 m/s.

What is collide?

Collide is a term used to describe the process of two objects or particles coming into contact with each other, often resulting in a collision. In physics, the term is used to refer to the force of two objects impacting one another. In everyday language, the term is used to describe two things, such as people or ideas, coming together in a way that produces a powerful impact.

The initial energy of the system can be calculated as:
[tex]E_{initial[/tex] = m₁*g*h + 0
where m_1 is the mass of the upper ball (2.0 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the vertical distance between the two balls (12.0 cm).
The final energy of the system can be calculated as:
[tex]E_{final} = (m_1 + m_2)\times v^2[/tex]
where m_1 and m_2 are the masses of the two balls (2.0 kg and 3.0 kg, respectively), and v is the velocity of the lower ball when the two balls stick together.
From these equations, we can solve for v:
[tex]v = sqrt[(m_1\timesg\timesh)/(m_1 + m_2)] = sqrt[(2.0 kg\times9.8 m/s^2\times12.0 cm)/(2.0 kg + 3.0 kg)] = 1.81 m/s[/tex]
Therefore, the velocity of the lower ball when the two balls stick together is 1.81 m/s.

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how large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius 150 m m at a speed of 121 km/h k m / h ?

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The coefficient of static friction, μs, between the tires and the road needs to be greater than the centripetal acceleration divided by the gravitational acceleration.

In this case, the centripetal acceleration can be calculated as ac = [tex](v^2)/r[/tex], where v is the speed and r is the radius of the curve. Therefore, the required coefficient of static friction μs = ac/g, where g is the gravitational acceleration, should be greater than μs = [tex](121 km/h)^2[/tex] / (150m) / [tex]9.81m/s^2[/tex] ≈ 0.93.


This means that the coefficient of static friction should be greater than 0.93 in order for the car to be able to round a level curve of radius 150 m at a speed of 121 km/h. This coefficient of static friction is necessary to counteract the centripetal force, allowing the car to round the curve without slipping.

If the coefficient of static friction is not large enough, the car will not be able to round the curve at the speed specified.

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4. if the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis, in what possible direction is the wave traveling?

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The possible direction in which an electromagnetic wave is traveling if the electric field is oscillating along the z-axis and the magnetic field is oscillating along the x-axis is the y-axis.

An electromagnetic wave is composed of two mutually perpendicular fields that oscillate perpendicular to the direction of the wave's propagation. They are the electric field and the magnetic field. An electromagnetic wave is created when a charged particle is accelerated. These waves can travel through a vacuum or any medium, including air and water, at the speed of light.

In this scenario, the electric field of the wave oscillates along the z-axis, while the magnetic field oscillates along the x-axis. As a result, the wave's propagation direction must be perpendicular to both fields. As a result, the wave must be propagating along the y-axis.This is why it's critical to comprehend the interplay between electric and magnetic fields in the context of electromagnetic waves.

It's also critical to recognize that an electromagnetic wave's direction of propagation is always perpendicular to the oscillation directions of the two fields, which are mutually perpendicular to each other.

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in which region are the temperatures and pressures at which it's possible to change the phase of x by raising or lowering the temperature?

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The region in which it is possible to change the phase of x by raising or lowering the temperature is: phase transition region.

This region is typically marked by an increase in pressure and a decrease in temperature. Temperature and pressure are inversely proportional to one another within this region, meaning that as pressure increases, temperature decreases and vice versa.

The exact temperature and pressure at which the phase transition occurs depends on the type of material being transitioned and its individual characteristics. For example, water boils at 100°C and 1 atm of pressure while other substances may have different boiling points.

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you compress a piston full of gas and do 8.4 joules of work on it. if the internal energy (u) of the system increases by 3.3 joules, how much heat (in joules) left the system (give your answer as a positive number)?

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The amount of heat that left the system is 11.7 joules (given as a positive number).

When a piston is compressed fully with gas and 8.4 joules of work is done on it, and the internal energy (u) of the system is increased by 3.3 joules, we need to determine the amount of heat that left the system.

To determine the amount of heat that left the system, we need to use the First Law of Thermodynamics, which states that the change in internal energy (u) of a system is the sum of the heat (q) added to it and the work (w) done on it, which can be represented as:

u = q + w

Where, u = Change in internal energy of the system

q = Heat added to the system

w = Work done on the system

From the given information, w = -8.4 J (since work was done on the system), and u = 3.3 J.

Therefore, substituting these values in the above equation, we get:

3.3 J = q + (-8.4 J)3.3 J + 8.4 J

q = 11.7 J

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a stone is thrown down off a bridge with a velocity of 5.6 m/s. what is its velocity after 3 seconds have passed?

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The velocity of the stone after 3 seconds have passed can be calculated using the formula v=u + at, where v is the velocity, u is the initial velocity, a is the acceleration (in this case the acceleration due to gravity, which is 9.8 m/s2), and t is the time. Therefore, the velocity of the stone after 3 seconds have passed will be 5.6 + (9.8*3) = 23.4 m/s.

The acceleration due to gravity causes any object to accelerate as it moves. This acceleration is always constant and acts downwards. Therefore, an object thrown with an initial velocity of 5.6 m/s will continue to accelerate and its velocity will increase. After 3 seconds have passed, the object will have an increased velocity of 23.4 m/s. In addition, when the stone is thrown off the bridge, it is subject to air resistance, which works against the stone and causes it to slow down. The magnitude of air resistance is dependent on a number of factors, such as the shape and size of the object. As such, the stone's velocity after 3 seconds might be slightly lower than the calculated value of 23.4 m/s.


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a 500g pot of water at room temperature (20c) is placed on a stove. how much heat is required to change this water to steam at 100c

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To change 500g of water at room temperature (20°C) to steam at 100°C, you will need to add 1128.500 kJ of heat. This is because water requires a certain amount of heat energy, called the 'latent heat of vaporization', to turn from a liquid to a gas.


Mass of water (m) = 500g

Initial temperature ([tex]T_i[/tex]) = 20°C

Final temperature ([tex]T_f[/tex]) = 100°C

The heat of vaporization ([tex]H_{vap}[/tex]) = 2260 J/g.

To calculate the amount of heat required to convert 500 g of water at room temperature to steam at 100°C, we will use the formula:

[tex]Q = m \times H_{vap}\\Q = 500 g \times 2260 J/g\\Q = 1128500 J[/tex]

Therefore, it would take 1130000 J of heat to change this water to steam at 100°C.

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The sound level produced by one singer is 71.8 dB. What would be the sound level produced by a chorus of 45 such singers (all singing at the same intensity at approximately the same distance as the original singer)? Answer in units of dB.

Answers

The sound level produced by a chorus of 45 singers would be approximately 88.3 dB.

How to find the sound level produced by a chorus of 45 singers?

Assuming that the sound level of each singer is independent and the same, the sound level produced by a chorus of 45 singers can be calculated using the following formula:

L2 = L1 + 10 log (N2/N1)

where:

L1 = the sound level of one singer = 71.8 dB

N1 = the number of singers in the original group = 1

N2 = the number of singers in the new group = 45

L2 = the sound level of the new group

Substituting the values in the formula, we get:

L2 = 71.8 + 10 log (45/1)

L2 = 71.8 + 10 log (45)

L2 = 71.8 + 16.5

L2 = 88.3 dB

Therefore, the sound level produced by a chorus of 45 singers would be approximately 88.3 dB, assuming all the singers are singing at the same intensity at approximately the same distance as the original singer.

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TRUE or FALSE – Energy can be transferred from Kinetic Energy (KE) to Potential Energy (PE) and vice versa.

Answers

True, energy can be transferred from kinetic energy (KE) to potential energy (PE) and vice versa

Can energy be transferred from Kinetic Energy (KE) to Potential Energy (PE) and vice versa?

The principle of the conservation of energy states that energy cannot be created or destroyed but can only transferred or transformed from one form to another.

When an object is in motion, it has kinetic energy, and when it is at rest, it has potential energy.

When the object moves from a stationary position to a position in motion, some of its potential energy is converted into kinetic energy.

Conversely, when the object moves from a position in motion to a stationary position, some of its kinetic energy is converted into potential energy.

Hence, the statement is TRUE.

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how much heat is lost through a 3' x 5' single-pane window with a storm that is exposed to a temperature differentia

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The amount of heat lost through a 3' x 5' single-pane window with a storm that is exposed to a temperature differential is 108 BTU per hour.

The U-factor is a measure of how well a window insulates against heat transfer. The lower the U-factor, the better the window insulates.

The temperature difference is the difference between the inside and outside temperatures.The area of the window is the size of the window.

Using these factors, we can calculate the rate of heat loss through the window in units of BTUs per hour.

Assuming a U-factor of 1.2 and a temperature difference of 60°F, the calculation would be:

Heat Loss = 1.2 BTU/(hrft^2F) x 15 ft^2 x 60°F

Heat Loss = 108 BTU/hour

Therefore, the heat lost through the window is 108 BTU per hour.

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Complete Question:

How much heat is lost through a 3' × 5' single-pane window with a storm that is exposed to a 60 Fahrenheit temperature differential?

suppose you have an atwood machine with two different masses m and m. what are the external forces acting on this system?

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The external forces acting on this system are: gravity and the tension in the string.

An Atwood machine is a system consisting of two masses, m, and m, connected by a string that passes over a pulley. In this system, the external forces are gravity and the tension in the string. Gravity pulls both masses downward, while the tension in the string acts in opposite directions on the two masses, pulling the heavier one down and the lighter one up.


The tension in the string is determined by the masses m and m and the acceleration of the system. If m is the heavier mass and m is the lighter mass, the tension in the string will be greater than if both masses had the same weight. This is because the tension must balance the gravitational forces on the two masses. The greater the mass, the greater the gravitational force, and the greater the tension in the string must be to balance it.


The acceleration of the system is determined by the masses, the tension in the string, and the amount of friction in the system. The greater the tension, the greater the acceleration, and the smaller the mass, the greater the acceleration. Friction acts against the acceleration, reducing the net acceleration of the system.


In summary, the external forces acting on an Atwood machine with two different masses m and m are gravity and the tension in the string. The tension in the string is determined by the masses and the acceleration of the system, while the acceleration is determined by the masses, the tension in the string, and the amount of friction in the system.

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g a cat with mass 4.50 kg is running at a speed of 6.70 m/s. what is the kinetic energy of the cat?

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The kinetic energy of the cat is 177.15 Joules.

The kinetic energy of the cat can be calculated using the formula K = 0.5mv2, where m is the mass and v is the velocity.

The cat has a mass of 4.50 kg and is running at a velocity of 6.70 m/s, so we can substitute these values into the formula to find the kinetic energy:

K = 0.5 * 4.50 kg * (6.70 m/s)2

K = 177.15 Joules

Kinetic energy is the energy possessed by an object due to its motion. It is calculated by multiplying half of the object's mass by its velocity squared.

The cat has a mass of 4.50 kg and is running at a velocity of 6.70 m/s, so its kinetic energy is 177.15 Joules.

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A long solenoid has 100 turns/cm and carries current i. an electron moves within the solenoid in a circle of radius 2.30 cm perpendicular to the solenoid axis. the speed of the electron is 0.0460c (c speed of light). find the current i in the solenoid.

Answers

The current in the solenoid becomes 3.56 A.

How to find current in the solenoid?

Number of turns in the solenoid, n = 100 turns/cm

Radius of the circular path of electron, r = 2.30 cm

Speed of electron, v = 0.0460c, where c is the speed of light

To find: Current in the solenoid, i

Formula used: Magnetic field inside the solenoid,

B = μ0ni Where, μ0 = 4π × 10⁻⁷ T m/A is the permeability of free spaceSolution:

The force on a moving electron in a magnetic field is given by

F = Bev

Where B is the magnetic field, e is the charge of an electron and v is its velocity.

The force acting on the electron provides the necessary centripetal force for the electron to move in a circle of radius r.

So,

Bev = (mev²)/r

where me is the mass of an electron

On simplifying the above equation, we get

Be = (mev)/r

Put the value of B from the formula of magnetic field inside the solenoid, B = μ0ni

we get

μ0ni = (mev)/r

Solve for i,

i = (mev)/(μ0nr)

Substitute the given values and solve

i = (9.109 × 10⁻³¹ kg × 0.0460c × 3 × 10⁸ m/s)/(4π × 10⁻⁷ T m/A × 100 turns/cm × 2.30 cm)i

= 3.56 A

Therefore, the current in the solenoid is 3.56 A.

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Hodan carried a box of (5,4)m. The box had a mass of 5kg. Hodan said that over 300J of work was done on the box. Is she correct, explain your answer​

Answers

Answer:

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a 170-hz sound travels through pure helium. the wavelength of the sound is measured to be 5.92 m. what is the speed of sound in helium?

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The speed of sound in pure helium is approximately 1006.4 m/s.

When a sound wave travels through a medium, it produces a series of compressions and rarefactions in the medium, which causes the particles of the medium to vibrate. The speed of sound in a particular medium depends on the physical properties of the medium, such as its density, elasticity, and temperature.

The speed of sound in helium can be calculated using the formula,

speed of sound = frequency x wavelength

Given that the frequency of the sound is 170 Hz and the wavelength is 5.92 m, we can plug in these values and get,

speed of sound = 170 Hz x 5.92 m

speed of sound = 1006.4 m/s

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what must the charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 700 n/c ?

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The charge (sign and magnitude) of a particle of mass 1.45 g must be for it to remain stationary when placed in a downward-directed electric field of magnitude 700 n/c is -1.029x10⁻⁴ C.

The magnitude of the charge must be equal to the magnitude of the electric field (700 n/c).

Therefore, we can write:-mg = qE

where, m = 1.45g = 1.45 x 10⁻³ kg

E = 700 N/cm = 1.45 x 10⁻³ kg x 9.81 m/s²

= 0.01419 N (Weight of the particle)

q = -1.029 x 10⁻⁴ C

To remain stationary when placed in a downward-directed electric field of magnitude 700 n/c, the charge (sign and magnitude) of a particle of mass 1.45 g must be negative.

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measurements show a certain star has a very high luminosity (100,000 x the sun's) while its temperature is quite cool (3500 k). how can this be?

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The star might be quite large in size, with a much larger surface area than the sun. This would increase its luminosity despite its cooler temperature.

The star has a high luminosity (100,000 x the sun's) and a cool temperature (3500 K) because of its size.

A star's luminosity is proportional to its size, so if a star is very large, it can have a high luminosity even if it is relatively cool.

Another possibility is that the star is in a phase of its life cycle where it has expanded and cooled, such as a red giant or supergiant, but still retains a high luminosity due to its large size.

These stars have relatively low surface temperatures, but their large sizes give them very high luminosities.

Therefore, this star is likely very large and thus has a very high luminosity despite its low temperature.

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