when rotating the platform, the hanging mass should be removed from the platform. question 2 options: true false

Work done on the can of soup by the attendant is 3J.

A supermarket checkout attendant can do a considerable amount of work on a can of soup when they apply a force of 5.00 N over a distance of 0.600 m.

Work is defined as the product of force and displacement and is calculated by multiplying the force applied (5.00 N) by the distance moved (0.600 m). Therefore, the work done on the can of soup is:

5.00 N * 0.600 m = 3.00 Nm.

To understand this concept further, it's important to know that the unit of work, joule (J), is equal to 1 Newton meter (Nm). So, the 3.00 Nm of work done on the can of soup is also equal to 3.00 J.

The amount of work done on the can of soup by the checkout attendant is directly proportional to the amount of force and displacement applied. If either of the two is increased, then the total work done will increase.

For example, if the checkout attendant applies a force of 10.00 N over a distance of 0.600 m, then the total work done will be 10.00 N * 0.600 m = 6.00 Nm, or 6.00 J.

In conclusion, a supermarket checkout attendant can do a considerable amount of work on a can of soup when they apply a force of 5.00 N over a distance of 0.600 m. This work is equal to 3.00 Nm or 3.00 J.

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The words on the pages of a textbook and the wave of a hand your friend makes when she sees you on the street are both examples of physical phenomena.

The words on the pages of a textbook and the wave of a hand your friend makes when she sees you on the street are both examples of physical phenomena. Physical phenomena are observable events or occurrences that can be described using the scientific method. These phenomena can be observed using our senses, such as sight, touch, sound, taste, and smell, or measured using instruments, such as thermometers, scales, or cameras. For example, the wave of a hand is a physical phenomenon because it is an observable event that can be seen and measured. Similarly, the words on the pages of a textbook are physical phenomena because they are observable and can be seen and read.

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The magnitude of the average rate in watts at which friction dissipates the box's mechanical energy is 10.4 W.

KE = 1/2 mv²

KE = 1/2 (5.00 kg) (2.50 m/s)² = 15.6 J

Since the box comes to rest after 1.50 s, the average power dissipated by friction can be found using the equation:

P = E/t

P = 15.6 J / 1.50 s = 10.4 W

Mechanical energy is the sum of the potential energy and kinetic energy of an object. Potential energy is the energy that an object possesses by virtue of its position or configuration in a field of force, while kinetic energy is the energy that an object possesses by virtue of its motion.

The mechanical energy of an object is conserved when it is in a closed system, meaning that the total amount of mechanical energy remains constant. This is known as the law of conservation of mechanical energy. For example, if an object is dropped from a height, its potential energy is converted into kinetic energy as it falls, and the total mechanical energy remains constant. Mechanical energy plays an important role in many areas of physics, including mechanics, thermodynamics, and electromagnetism.

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Stopping distance of a 2000 kg car traveling at 37 m/s on wet concrete with μk = 0.50 is 141.95 meters.

To solve this problem, we need to use the formula for the stopping distance of a car on a slippery surface:

d = (v² / 2μk g)

where:

v - is the car's initial velocity.

d - is the stopping distance

μk - is the coefficient of kinetic friction between the car's tires and the road surface

g - is the acceleration due to gravity (9.81 m/s²)

Substituting the given values, we get:

d = (37² / (2 * 0.50 * 9.81)) = 141.95 meters

Therefore, the stopping distance of the car is 141.95 meters.

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Complete question:

A 2000 kg car traveling at a speed of 37 m/s skids to a halt on wet concrete where μk = 0.50. What is Stopping distance?

The x-component of the magnetic field produced by these three wires at the origin, Bx(0,0), is equal to zero since the wires are aligned parallel to the z-axis.

The y-component of the magnetic field produced by these three wires at the origin, By(0,0), is equal to μ0/2π times the sum of the currents, or 0.8 x 10-7 T.

The x-component of the force exerted on a one meter length of the wire carrying current I1, Fx(1), is equal to the product of the current, I1, and the y-component of the magnetic field, By(0,0). This is equal to -2.7 x 10-7 N.

The y-component of the force exerted on a one meter length of the wire carrying current I1, Fy(1), is equal to the product of the current, I1, and the x-component of the magnetic field, Bx(0,0). Since Bx(0,0) is equal to zero, the force is equal to zero.

The x-component of the force exerted on a one meter length of the wire carrying current I2, Fx(2), is equal to the product of the current, I2, and the y-component of the magnetic field, By(0,0). This is equal to 0.4 x 10-7 N.

In conclusion, the x-component of the magnetic field produced by these three wires at the origin, Bx(0,0), is equal to zero. The y-component of the magnetic field produced by these three wires at the origin, By(0,0), is equal to 0.8 x 10-7 T.

The x-component of the force exerted on a one meter length of the wire carrying current I1, Fx(1), is equal to -2.7 x 10-7 N. The y-component of the force exerted on a one meter length of the wire carrying current I1, Fy(1), is equal to zero. The x-component of the force exerted on a one meter length of the wire carrying current I2, Fx(2), is equal to 0.4 x 10-7 N.

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The height of the image is approximately 1.066 cm, and since it is negative, it means that the image is inverted and smaller than the object.

Using the thin lens equation:

1/f = 1/d_o + 1/d_i

where f is the focal length of the lens, d_o is the object distance, and d_i is the image distance.

Plugging in the given values, we get:

1/4 = 1/16 + 1/d_i

Solving for d_i, we get:

d_i = 3.2 cm

Using the magnification equation:

m = -d_i/d_o

where m is the magnification of the image.

Plugging in the given values, we get:

m = -3.2/16 = -0.2

Since the magnification is negative, the image is inverted.

Finally, using the equation:

m = h_i/h_o

where h_i is the height of the image, and h_o is the height of the object.

Plugging in the given values and solving for h_i, we get:

h_i = m * h_o = (-0.2) * 5.33 cm = -1.066 cm

Therefore, the height of the image is approximately 1.066 cm, and since it is negative, it means that the image is inverted and smaller than the object.

What is magnification of lens?

The magnification of a lens is a measure of how much larger or smaller an image appears relative to the object that is being viewed through the lens. It is the ratio of the height of the image formed by the lens to the height of the object.

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The rock will be 11 meters from the ground 2.97 seconds after it is thrown.

Let's start by using the kinematic equation,

h = vit + 0.5a*t^2

where h is the height, vi is the initial velocity, t is the time, and a is the acceleration due to gravity (-9.8 m/s^2).

At the highest point, the rock's velocity will be zero, so we can use this fact to find the time it takes to reach the highest point,

0 = 13 - 9.8*t_highest

t_highest = 1.33 seconds

Now we can use this time to find the height of the rock above the ground,

h = 38 + 131.33 - 0.59.8*(1.33)^2

h = 51.33 meters

So at its maximum height, the rock is 51.33 meters above the ground. To find when it will be 11 meters from the ground,

11 = 51.33 + 0 + 0.5*(-9.8)*t^2

t^2 = (51.33 - 11)/4.9

t^2 = 8.8367

t = 2.97 seconds (rounded to two decimal places)

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Zero resistance or capacitance value corresponds to a short circuit, which is the travel of current along an unintended path.

True short circuits happen when electrical circuit wires or wire connections are exposed or broken; they need to be identified and addressed as soon as possible. When there is a low resistance connection between two conductors supplying electricity to a circuit, a short circuit happens.

A "ideal" open circuit would have zero capacitance. A capacitor with 0 capacitance has no electrical charge accumulating on its plates or conductors. Zero capacitance means it can become fully charged as soon as the current is flown through it.

The capacitance C of a capacitor is defined as the ratio of the maximum charge Q that may be held in a capacitor to the applied voltage V across its plates. In other terms, capacitance is the capacity of the device to store the most charge per volt:

C = Q/V.

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D. Two substances must combine chemically.

One type of atom makes up an element, which is a white crystalline solid that cannot be divided into more than one component. Compounds are pure substances created by mixing two or more substances in a specific mass ratio.

Chemically pure substances that can be found in nature are called elements and compounds. An component is an object made of the same kinds of atoms as a compound, which is how they differ from one another.

It is common for the disorder to develop and the entropy of the system to increase as a result of the dispersion of molecules, atoms, and ions of one sort during the creation of a solution throughout a second substance.

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A 5-gallon bucket of gold would weigh 86.84 pounds.

A five-gallon bucket of water weighs 40 pounds. Gold has a specific gravity of almost 20.

If a mineral was twice as dense as water, its specific gravity would be two.

Water has a specific gravity of 1.

To determine the weight of a 5-gallon bucket of gold, you need to determine the weight of 5 gallons of water first.One gallon of water weighs approximately 8.33 pounds; hence 5 gallons of water weigh 41.65 pounds.

Now, divide the weight of 5 gallons of water (41.65) by the specific gravity of gold (20):41.65/20 = 2.0825

The weight of a five-gallon bucket of gold would be 2.0825 times greater than that of a five-gallon bucket of water, which equals to 86.84 pounds (40 pounds + 46.84 pounds).

Therefore, a 5-gallon bucket of gold would weigh approximately 86.84 pounds.

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Answer:

Typical golf caddie responsibilities include carrying clubs, cleaning balls,

calculating distances and scores, and even replacing the divots - pieces of

grass and dirt that have been cut loose by the swift, forceful strikes of golf

clubs.

The final answer are velocity of the object when it is midway to the ground is : v  = u + at v = 0 + 9.81 x 0.78v = 7.658 m/s= 8.77 m/s (rounded to two decimal places)

At the midpoint, the velocity of the 4.5 kg item dropped from a height of 6.0 m is 8.77 m/s (use g = 9.81 m/s^2).

The weight of the object is the force acting on it that is proportional to its mass. The weight is represented by the formula W=mg. W = 4.5 x 9.81 m/s²W = 44.145 N, where N is the newton.

The object's initial velocity is zero, and it starts to fall. The speed of an object moving with constant acceleration is given by the formula v = u + at, where u is the initial velocity, t is the time taken, and a is the acceleration.

In this instance, the initial velocity u = 0. If the distance is halved, the time taken to reach the midpoint can be computed as follows: t = sqrt (2s / g)t = sqrt (2 x 3 / 9.81)t = sqrt (0.611) t = 0.78 s

The velocity of the object when it is midway to the ground is : v  = u + at v = 0 + 9.81 x 0.78v = 7.658 m/s= 8.77 m/s (rounded to two decimal places)

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The two charges would be 219.1 meters apart.

The force between two charges is given by Coulomb's Law:

F = k * (q1 * q2) / r^2

where F is the force, k is the Coulomb constant (9.0 x 10^9 N*m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.

In this case, we have two charges with a magnitude of +2.0 C each and a force of 8.5 x 10^8 N between them. Substituting these values into Coulomb's Law, we get:

8.5 x 10^8 N = (9.0 x 10^9 N*m^2/C^2) * [(2.0 C)^2 / r^2]

Simplifying this expression, we get:

r^2 = [(2.0 C)^2 * (9.0 x 10^9 N*m^2/C^2)] / (8.5 x 10^8 N)

r^2 = 4.8 x 10^4 m^2

Taking the square root of both sides, we get:

r = 219.1 meters

Therefore, the two charges are 219.1 meters apart.

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The 5-volt reading we can expect between the two test points if the reference lead is on the lowest voltage.

The given data is as follows:

The first test marked voltage = 5 volts

The second test marked voltage = -3.3 volts

Let us assume that the two test points are there is a conductive track between them, the voltage between the two points can be calculated using the voltage difference between the two test points.

The voltage difference between the  two test points is calculated as:

5 volts - (-3.3 volts) = 8.3 volts

If the reference lead is on the lowest voltage, It means that the negative side of the voltmeter is attached to the test point with the lower voltage which is -3.3 volts.

The voltage difference between the  two test points is

8.3 volts - 3.3 volts = 5 volts

Therefore we can conclude that the 5-volt reading we can expect between the two test points.

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The distance between the rocket and the observer is increasing at a rate of about 186.6 m/s.

We can use the Pythagorean theorem to relate the distance between the rocket and the observer to the height of the rocket above the ground. Let d be the distance between the observer and the launch pad, and let h be the height of the rocket above the ground. Then,

d^2 = h^2 + 3^2 (1)

We can take the derivative of both sides of equation (1) with respect to time to get,

2d (dd/dt) = 2h (dh/dt) (2)

where (dd/dt) is the rate of change of distance between the observer and the rocket, and (dh/dt) is the rate of change of height of the rocket.

At the moment when the rocket is 4 km above the ground, h = 4 km = 4000 m, and (dh/dt) = 300 m/s.

Substituting these values into equation (2) and solving for (dd/dt),

dd/dt = (h/d) x (dh/dt) = (4000 m / √(4000^2 + 3000^2) m) x (300 m/s)

≈ 186.6 m/s

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Nuclear reactor incidents have been studied in hundreds to determine the stochastic effects on the workers and exposed population.

The stochastic effect refers to radiation-induced effects that may occur in tissues or cells and that are subject to probabilistic relationships between exposure and reaction. The probability of developing cancer increases as a result of exposure to radiation. The greater the exposure dose, the greater the likelihood of developing cancer.The stochastic effect can occur even at low radiation doses.

This is opposed to deterministic effects, which only occur when a particular radiation dose threshold is surpassed. Stochastic effects are also referred to as random or probabilistic effects. They can happen in any tissue or organ in the body, including reproductive cells, which can lead to heritable genetic mutations. The probability of developing cancer increases as the radiation dose rises.

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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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All of the following have wavelengths that are longer than visible light: Gamma Rays, Ultraviolet (UV) Light, Infrared Radiation, and X Rays (A, B, and D).

Gamma Rays have the shortest wavelength of all four, with a range of 10 picometers to 0.01 nanometers. Ultraviolet (UV) Light has a range of 10 nanometers to 400 nanometers. Infrared Radiation has a range of 700 nanometers to 1 millimeter. Finally, X Rays have a range of 0.01 nanometers to 10 nanometers.
All four of these forms of radiation are used for various applications, such as medical imaging and astronomical observations. Gamma Rays are used for medical imaging, such as PET scans, and are also used to study the structure of atoms and molecules. Ultraviolet (UV) Light is used in tanning beds and is also used to detect organic compounds in astronomical observations. Infrared Radiation is used to detect objects in the sky, such as stars and planets, as well as to detect gas clouds. Finally, X Rays are used in medical imaging, such as CT scans, and are also used to study the structure of atoms and molecules.  

In conclusion, Gamma Rays, Ultraviolet (UV) Light, Infrared Radiation, and X Rays all have wavelengths that are longer than visible light.

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Based on the above,  3.75 x 10¹² electrons have been added to the glass rod.

To determine the number of electrons added to the glass rod, we need to know the charge of a single electron. One electron has a charge of -1.6 x 10⁻¹⁹ C.

Charge added to the glass rod = -0.6 μC = -0.6 x 10-⁶ C

Number of electrons added to the glass rod = (charge added to the rod) / (charge of a single electron)

Number of electrons added to the glass rod = (-0.6 x 10⁻⁶ C) / (-1.6 x 10⁻¹⁹C) = 3.75 x 10¹² electrons

Therefore, 3.75 x 10¹² electrons have been added to the glass rod.

For the second question, we can use Coulomb's law to determine whether the pith ball and metal plate are attracted or repulsed. Coulomb's law states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them where

The force between the pith ball and the metal plate is:

F = k x q1 x  q2 / r²

Pugging the value in the formula, the answer will be:  F = -0.135 N.

The negative sign indicates that the force is attractive, so the pith ball and metal plate are attracted to each other.

For the third question, the force on the negatively charged object from problem 2 is 0.135 N, in the direction towards the positively charged pith ball.

For the fourth question, the force on the positively charged object from problem 2 is also 0.135 N, in the direction towards the negatively charged metal plate.

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See text below

You rub a glass rod with a piece of fur. If the rod now has a charge of -0.6 μC, how many electrons have been added to the rod?

A suspended pith ball possessing +10 μC of charge is placed 0.02 m away from a metal 'plate possessing -6 μC of charge. Are these objects attracted or repulsed?

What is the force on the negatively charged object from problem 2?

• What is the force on the positively charged object from problem 2?

The force exerted by the boy on the man is also 100 N, according to Newton's Third Law of Motion.

Newton's Third Law of Motion is a fundamental principle in physics that describes the relationship between forces acting on objects. It states that for every action, there is an equal and opposite reaction.

This means that when an object exerts a force on another object, the second object exerts an equal and opposite force back on the first object.

In this scenario, the man exerts a force of 100 N on the boy in the forward direction. As a result, the boy exerts an equal and opposite force of 100 N on the man in the backward direction.

This principle applies to all objects in motion and is essential for understanding the dynamics of physical systems.

Therefore, the force exerted by the boy on the man is also 100 N.

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The rate of change of the boulder's momentum is equal to the rate of change of the pebble's momentum. Option C is correct.

This is because the rate of change of an object's momentum is directly proportional to the net force acting on the object, and inversely proportional to the mass of the object. In this case, the net force acting on both the boulder and the pebble is the same, at 200 N. However, the mass of the boulder is much larger than the mass of the pebble.

Since the rate of change of momentum is inversely proportional to the mass of the object, the boulder will experience a smaller rate of change in momentum than the pebble. However, this will be exactly offset by the fact that the boulder has a larger mass, which will cause its momentum to change at the same rate as the pebble.

Therefore, the rate of change of the boulder's momentum is equal to the rate of change of the pebble's momentum, despite the large difference in their masses. The principle behind the rate of change of momentum is that the amount of momentum an object has is directly proportional to its mass and velocity. When a net force acts on an object, it causes the object's velocity to change, which in turn causes a change in the object's momentum.

The rate of change of an object's momentum is determined by the net force acting on the object, as well as its mass. Specifically, the rate of change of momentum is equal to the net force acting on the object divided by its mass. This principle is known as Newton's second law of motion.

In the case of the boulder and the pebble in the original question, both objects are subject to the same net force of 200 N. However, the boulder has a mass of 100 kg, while the pebble has a mass of 0.13 kg. Since the rate of change of momentum is inversely proportional to the mass of the object, the pebble will experience a much larger rate of change in momentum than the boulder. Option C is correct.

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A particle travels 17 times around a 16-cm radius circle in 38 seconds.The average speed of the particle is approximately 14.21 m/s.

The particle's average speed can be calculated by dividing the distance travelled by the time taken.

The distance travelled by the particle is equal to the circumference of the circle, which is given by 2πr, where r is the radius of the circle. Therefore, the distance travelled by the particle is given by:

Distance travelled = 17 × 2π × 16 cm

                               = 17 × 32π cm

The time taken by the particle to travel this distance is given as 38 seconds.Therefore, the average speed of the particle is given by:

Average speed = Distance travelled /Time taken

                          = (17 × 32π) cm/38 s

                          = 17 × 32 × π/38 m/s

                          = 14.21 m/s (rounded to two decimal places).

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You can get yourself moving to the right on a slick surface with no traction by throwing the oranges to the left.

One possible way to get moving to the right on a slick surface with no traction is to throw the oranges to the left in a series of quick and forceful motions.

According to Newton's third law of motion, every action has an equal and opposite reaction. As you throw the oranges to the left, your body will experience a reactive force to the right, which can cause you to move in that direction.

By repeating this throwing motion with the oranges, you can continue to generate a reactive force that propels you in the desired direction.

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Hence, 127.2 m3/s per second is the required water flow rate from the dam's crest.

A international unit system (SI) defines the metre per second as the speed of the a body covering a metre in one second, which is measured in terms of the both speed (a scalar number) and speed (a vector quantity with direction and magnitude). m/s, m/s1, m/s, or ms are the SI unit symbols.

Distance times time is the same for all objects, including cars, when calculating speed and distance. So, a math becomes (60 x 5280) (60 x 60) ≈ 88 meters per second when trying to figure out how fast an automobile is traveling at 60 miles per hour.

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The resistance of a wire is quantified by r. The wire is stretched to double its length, but the material maintains the same density.

When a wire's length is doubled, its resistance is doubled as well. When a wire's thickness is doubled, its resistance is decreased to one-half of its previous value. The formula for wire resistance is given by the following equation:

R = ρL / A

Since the wire's density is constant, the resistance is proportional to the wire's length and inverse to the cross-sectional area. As a result, doubling the wire's length would double its resistance.

So the new resistance in r would be twice the previous resistance, which is R.

Mathematically,

[tex]R_2[/tex] = 2R where R is the previous resistance of the wire.

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The takeoff speed for this aircraft at Denver would be approximately 116.85 moh if the speed of takeoff of the aircraft at sea level is 120 moh.

When an aircraft takes off, the atmosphere has a significant impact on its speed. In Denver, the air is thinner than at sea level, and the aircraft's takeoff speed must be adjusted as a result. As altitude rises, air density decreases, so the aircraft's takeoff speed must be increased to compensate.The formula for calculating takeoff speed with respect to altitude is given below:

Takeoff speed at altitude h = Takeoff speed at sea level x √(air density at altitude h / air density at sea level)

We know that the takeoff speed at sea level is 120 moh. Let us assume that air density at Denver is 0.91 times the air density at sea level.Hence, the takeoff speed at Denver can be calculated as:

Takeoff speed at Denver = 120 x √(0.91)≈ 116.85 moh.

Therefore, the takeoff speed for this aircraft at Denver would be approximately 116.85 moh.

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Option C, It would take 50 m³ of Mars' atmosphere to collect the same mass of air as one m³ on Earth. To calculate the volume of Mars' atmosphere required to collect a mass of 1kg, we need to use the density of the Martian atmosphere and the mass of the air on Earth.

The density of air at moderate altitude on Earth is given as 1 kg/m3. This means that 1 cubic meter of air on Earth has a mass of 1 kg. To convert this to grams per cubic centimeter, we can divide by 1000, which gives 0.001 g/cm3.

The mass of air in one m³ on Earth is 1 kg, while the density of the atmosphere near Mars' surface is 0.02 kg/m³. Therefore, to collect 1 kg of Mars' atmosphere, we need:

1 kg / 0.02 kg/m³ = 50 m³

So, it would take 50 m³ of Mars' atmosphere to collect the same mass of air as one m³ on Earth.

Complete question -

The density of air at moderate altitude on earth is 1 kg/m3 (this can be converted to 0.001 g/cm3). the density of the atmosphere near mars' surface is 0.02 kg/m3. how many m3 of mars atmosphere would it take to collect a mass of 1kg, the same mass as in one m3 on earth?

A. 1

B. 10

C. 50

D. 100

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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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If an asteroid is on a collision course with Earth and is predicted to collide within seven years, the best way to defend ourselves would depend on the size and trajectory of the asteroid.

An asteroid is a small, rocky object that orbits the Sun. Most asteroids are found in the asteroid belt, a region between the orbits of Mars and Jupiter. Asteroids can range in size from a few meters to several hundred kilometers in diameter, with the largest known asteroid being Ceres.

Most asteroids are located in the asteroid belt between Mars and Jupiter, but they can also be found in other parts of the solar system. Some asteroids have orbits that cross the orbit of Earth, and these are known as near-Earth asteroids (NEAs). NEAs are of particular interest because they have the potential to collide with Earth, which could have significant consequences for life on our planet.

Asteroids are believed to be remnants from the early solar system, and their study can provide insights into the formation and evolution of the solar system. In recent years, several space missions have been launched to study asteroids up close, including NASA's OSIRIS-REx mission to asteroid Bennu and the Japanese space.

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