Un bouteille de jus d’orange contient 100 g de sucre. Une cannette de 33 cL de soda contient 34 g de sucre.
1. Quelle est la concentration en sucre dans un verre de 20 cL de jus d’orange ? 2. Quelle est la masse de sucre dans un verre de 20 cL de jus d’orange ?
3. Comparer les concentration en sucre du soda et du jus d’orange.

The car travels approximately 245 m horizontally and 14.1 m vertically in 15 seconds up an inclined hill.

Section A: To find how far evenly the vehicle has gone in 15 s, we really want to utilize the equation:distance = starting speed × time + 1/2 × speed increase × time^2.Since the vehicle is beginning from rest, the underlying speed is 0. We are searching for the even distance, which is the distance along the ground. Hence, we can disregard the grade of the slope. Hence, we can utilize the level part of the speed increase, which is given by:a_horizontal = a × cosθ.where θ is the point of grade and an is the speed increase. Subbing the qualities given, we get:a_horizontal = 2.2 m/s^2 × cos(5.3°) ≈ 2.18 m/s^2.Utilizing this worth, we can find the even distance went in 15 s:distance = 0 + 1/2 × 2.18 m/s^2 × (15 s)^2 ≈ 245 m.Hence, the vehicle has voyaged around 245 m evenly in 15 s.

Part B: To find how far in an upward direction the vehicle has gone in 15 s, we want to utilize the upward part of the speed increase, which is given by:a_vertical = a × sinθ

Subbing the qualities given, we get:

a_vertical = 2.2 m/s^2 × sin(5.3°) ≈ 0.197 m/s^2

Utilizing this worth, we can find the upward distance went in 15 s:

distance = 0 + 1/2 × 0.197 m/s^2 × (15 s)^2 ≈ 14.1 m

Hence, the vehicle has voyaged roughly 14.1 m in an upward direction in 15 s.

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The correct answer for all questions are: a), b) & c) will be acceleration due to gravity g=9.8 m/s^2, d) 0 m/s, e) 7.644 m/s, f) 2.977 m

A) The acceleration of the ball while it is moving upward is the acceleration due to gravity, which is 9.8 m/s2 in the downward direction.

B) The acceleration of the ball while it is moving downward is also the acceleration due to gravity, which is 9.8 m/s2 in the downward direction.

C) The acceleration of the ball while it is at its maximum height is still the acceleration due to gravity, which is 9.8 m/s2 in the downward direction.

D) The velocity of the ball when it reaches its maximum height is 0 m/s, because it has stopped moving upward and is about to start moving downward.

E) The initial velocity of the ball can be found using the equation v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time.

Since the ball is caught at the same height it was released, the final velocity is equal to the initial velocity, but in the opposite direction. Therefore, v = -v0. Plugging in the values for a (9.8 m/s2) and t (1.56 s), we get:

-v0 = v0 + (9.8 m/s2)(1.56 s)

Solving for v0, we get:

v0 = (9.8 m/s2)(1.56 s)/2

v0 = 7.644 m/s

The initial velocity of the ball is 7.644 m/s in the upward direction.

F) The maximum height that the ball reaches can be found using the equation h = v0t + (1/2)at2, where h is the height, v0 is the initial velocity, t is the time, and a is the acceleration.

Plugging in the values for v0 (7.644 m/s), t (1.56 s/2 = 0.78 s), and a (9.8 m/s2), we get:

h = (7.644 m/s)(0.78 s) + (1/2)(9.8 m/s2)(0.78 s)2

h = 2.977 m

The maximum height that the ball reaches is 2.977 m.

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Answer:

[tex]\displaystyle 2\, \pi\, \sqrt{\frac{R^{3}}{G\m M}}[/tex], where [tex]R[/tex] is the orbital radius, [tex]M[/tex] is the mass of the Moon, and [tex]G[/tex] is the gravitational constant.

Explanation:

Let [tex]m[/tex] denote the mass of the satellite. Let [tex]R[/tex] denote the orbital radius, let [tex]M[/tex] denote the mass of the Moon, and let [tex]G[/tex] denote the gravitational constant.

The Moon would exert the following gravitational attraction on the satellite:

[tex]\displaystyle \frac{G\, M\, m}{R^{2}}[/tex].

Let [tex]\omega[/tex] denote the angular velocity of the satellite. For the satellite to stay in this orbit of radius [tex]R[/tex], the net force on the satellite needs to be:

[tex]m\, \omega^{2}\, R[/tex].

Since the gravitational force is the only force on this satellite, the net force on the satellite would be equal to the gravitational force:

[tex]\displaystyle m\, \omega^{2}\, R = \frac{G\, M\, m}{R^{2}}[/tex].

Rearrange this equation to find the angular velocity:

[tex]\displaystyle \omega^{2} = \frac{G\, M}{R^{3}}[/tex].

[tex]\displaystyle \omega = \sqrt{\frac{G\, M}{R^{3}}}[/tex].

Note that with the Moon as the center, a full revolution around the Moon would take an angular distance of [tex]2\, \pi[/tex]. Divide the angular distance by the angular velocity to find the time required for this revolution:

[tex]\begin{aligned}T &= \frac{2\, \pi }{\omega} && \genfrac{}{}{0}{}{\text{angular displacement}}{\text{angular velocity}} \\ &= 2\, \pi \, \sqrt{\frac{R^{3}}{G\, M}}\end{aligned}[/tex].

The power consumed by 18 watts.

What is power ?

Work is done at a certain rate according to power. P=Wt. The watt (W), which is equal to 1 joule per second (1J/s), is the SI unit of power. Power is the rate at which energy is used up since power is the transfer of energy. For instance, a 60-W light bulb uses about 60 J of energy every second.

What is current ?

A flow of electrical charge carriers, most frequently electrons or atoms lacking in electrons, is referred to as current. I in capital letters is a typical symbol for current. Ampere, which is denoted by the letter A, is the common unit.I, which is derived from the French term intensité du courant, is the commonly used symbol for current (current intensity).

Therefore,  power consumed by 18 watts.

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4.43 seconds

The length of the pendulum and the acceleration caused by gravity affect how long an object will oscillate when it is suspended from the ceiling by a 16-foot string.

The following formula determines a pendulum's period T:

T = 2π * sqrt(L/g)

where g is the acceleration caused by gravity and L is the pendulum's length.

The pendulum in this scenario is 16 feet long. The acceleration brought on by gravity is roughly 32 feet per second. As a result, the oscillation period T is:

T = 2π * sqrt(16/32) = 2π * sqrt(1/2) = 2π * 0.707 = 4.43 seconds (approximately)

T = 2π * sqrt(16/32)

  = 2π * sqrt(1/2)

  = 2π * 0.707  

  = 4.43 seconds (approximately)

Thus, the object's period of oscillation is approximately 4.43 seconds.

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The speed of the edge of a fidget spinner that has a diameter of 6.0 cm is 12.58 m/s.

To calculate the speed of the edge of the fidget spinner, we need to convert the rotational speed from revolutions per minute to radians per second, and then multiply by the radius of the spinner.

First, we convert 4000 rpm to radians per second:

4000 rpm = 4000 revolutions / 60 seconds = 66.67 revolutions per second

Each revolution is 2π radians, so:

66.67 revolutions per second = 66.67 * 2π radians per second = 419.47 radians per second

Now that we have the rotational speed in radians per second, we can calculate the speed of the edge of the fidget spinner:

speed = rotational speed x radius

The radius of the fidget spinner is half of its diameter, which is 6.0 cm / 2 = 3.0 cm = 0.03 m.

So, the speed of the edge of the fidget spinner is:

speed = 419.47 radians per second x 0.03 m = 12.58 m/s

Therefore, the speed of the edge of the fidget spinner is approximately 12.58 m/s.

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A sample of helium gas occupies 5 liters at 22°C and a gauge pressure of 200,000 pascals. The mass of the gas is 0.16 gm. It is calculated using ideal gas equation.

Define gauge pressure.

When compared to atmospheric pressure, gauge pressure is the pressure that is higher; it is positive for pressures over atmospheric pressure and negative for pressures below atmospheric pressure. Every fluid that is not contained experiences additional pressure due to the atmospheric pressure.

Gauge pressure, which is equal to the difference between absolute and atmospheric pressure and is zero-referenced against ambient air pressure, is the additional pressure in any system relative to atmospheric pressure. The absence of the negative sign serves to distinguish the negative pressure and is typically ignored.

The word "vacuum" can be given a value, and the gauge is referred to as a vacuum gauge. When internal pressure exceeds atmospheric pressure in a given area, the phrase gauge pressure is employed.

Using the formula:

PV = nRT

substituting the values and solving for n:

n = 0.04 moles

mass = 0.04 × 4 = 0.16g

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D, invisible is not a way light can be transmitted.

Light can be transmitted through a medium such as air, water, glass, or a vacuum. When light is transmitted, it passes through the medium without being absorbed or reflected, allowing it to be seen on the other side of the medium. Transmission of light can occur in a straight line or can be refracted or diffracted, depending on the properties of the medium it is passing through.

Light can be absorbed by a material, which means that the energy of the light is transferred to the material. When this happens, the material may become excited, and the energy can cause a change in the material, such as heating it up, causing it to emit light, or causing a chemical reaction.

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If the basketball has a radius of 12cm,the buoyant force on the basketball due to the water is 35.5 N.

When an object is submerged in a fluid, it experiences an upward force called the buoyant force. The magnitude of the buoyant force is equal to the weight of the fluid displaced by the object.

In this case, the basketball is submerged in water until half its volume is submerged. The volume of a sphere of radius r is given by:

V = (4/3) * pi * r^3

So, the volume of the basketball is:

V_basketball = (4/3) * pi * (0.12 m)^3 = 7.24 x 10^-3 m^3

Half of the volume of the basketball that is submerged is:

V_submerged = (1/2) * V_basketball = 3.62 x 10^-3 m^3

The weight of the water displaced by the basketball is equal to the weight of the volume of water that is submerged. The density of water is 1000 kg/m^3, so the mass of the displaced water is:

m_water = rho * V_submerged = 1000 kg/m^3 * 3.62 x 10^-3 m^3 = 3.62 kg

The buoyant force on the basketball due to the water is equal to the weight of the displaced water, which is:

F_buoyant = m_water * g

= 3.62 kg * 9.81 m/s^2

= 35.5 N

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Answer: Option : C and D are correct .

Explanation: Because core increases the strength of the magnetic field since core is having iron and nickel metals that helps to increase the current.

The magnitude and direction of the minimum magnetic field needed to levitate the rod is 0.152 T.

The magnetic force on a current-carrying wire in a magnetic field is given by:

F = BIL sin(θ)

where B is the magnetic field,

I is the current in the wire,

L is the length of the wire in the magnetic field, and

θ is the angle between the direction of the magnetic field and the direction of the current.

Therefore, we want to find the minimum magnetic field required to levitate the copper rod the rod is levitating, the magnetic force acting upwards must be equal and opposite to the gravitational force acting downwards.

The gravitational force on the rod is given by:

F_{gravity }= mg

where m is the mass of the rod and

g is the acceleration due to gravity.

Substituting the given values, we get:

F_{gravity} = (0.043 kg) * (9.81 m/s²) = 0.42283 N

For the rod to levitate, the magnetic force must be equal in magnitude and opposite in direction to the gravitational force:

F_{magnetic} = F_{gravity}

Substituting the expression for the magnetic force, we get:

BIL sin(θ) = mg

Solving for B, we get:

B = mg / IL sin(θ)

Substituting the given values, we get:

B = (0.043 kg * 9.81 m/s²) / (15 A * 0.61 m * sin(90°))

B = 0.152 T

Therefore, the magnitude of the minimum magnetic field needed to levitate the copper rod is 0.152 T. The direction of the magnetic field should be perpendicular to the direction of the current flow in the rod, which is in the positive x direction in this case.

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The total angular momentum of the two-disk system after the smaller disk is dropped on the larger disk is [tex]\omega _2 = (I_1\times \omega _1) / (I_1 + m_2 \times r^2)[/tex].

The total angular momentum of the two-disk system after the smaller disk is dropped on the larger one will depend on the initial angular momenta of the two disks and any changes that occur when they collide.

Assuming the system is isolated, the total angular momentum before and after the collision must be conserved. Before the collision, we can assume that the smaller disk has an initial angular momentum of zero since it is not rotating, and the larger disk has an initial angular momentum given by:

[tex]L_1 = I_1 \times \omega_1[/tex]

where [tex]I_1[/tex] is the moment of inertia of the larger disk and ω₁ is its initial angular velocity.

When the smaller disk is dropped onto the larger one, there will be a transfer of angular momentum from the smaller disk to the larger one. The final angular velocity of the combined system will depend on the moment of inertia of the combined system, which can be approximated as:

[tex]I_2  I_1 + m_2 \times r^2[/tex]

Where m_2 is the mass of the smaller disk and r is the distance from the center of the larger disk to the point where the smaller disk makes contact. We assume that the collision is elastic and that no external forces act on the system, so angular momentum is conserved:

[tex]L_1 = L_2[/tex]

[tex]I_1 \times \omega _1 = (I_1 + m_2 \times r^2) \times \omega _2[/tex]

Solving for ω₂, we get:

[tex]\omega _2 = (I_1\times \omega _1) / (I_1 + m_2 \times r^2)[/tex]

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Contact metamorphism is least likely to occur in low-temperature areas with sedimentary rocks, such as the Earth's oceanic crust.

Contact metamorphism occurs due to heating, with or without burial, of rocks that lie close to a magma intrusion. It is characterized by low P/T gradients, as strong thermal gradients between an intruding magma and adjacent country rock are best established at shallow crustal level. It is a type of metamorphism where rock minerals and texture are changed, mainly by heat, due to contact with magma. Regional metamorphism is a type of metamorphism where rock minerals and texture are changed by heat and pressure over a wide area or region.

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The work done by the capacitor is obtained as  1.36 * 10^-5 J

A capacitor is an electrical component that stores electrical energy in an electric field. It consists of two conductive plates separated by a non-conductive material, known as the dielectric.

When a voltage is applied across the plates, one plate becomes positively charged and the other becomes negatively charged, creating an electric field between them. The dielectric material prevents the charges from flowing between the plates and instead stores the energy in the form of an electric field.

Work done by the Capacitor = 1/2CV^2

= 0.5 * 2.72 * 10^-9 * (100)^2

= 1.36 * 10^-5 J

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Missing parts;

A 2. 72-nF parallel-plate capacitor is charged to an initial potential difference ΔVi = 100 V and is then isolated. The dielectric material between the plates is Pyrex glass, with a dielectric constant of 5. 6. What is the work done by the capacitor?

When a luminous object moves along the optical axis of a concave spherical mirror, if the object approaches the mirror and its focal point, the image of the object moves to infinity.

A concave mirror is a mirror with a reflecting surface that curves inward. A concave mirror, often known as a converging mirror, causes light to concentrate at a point known as the focal point when light rays hit it. The distance between the mirror's center and its focal point is known as the focal length. A concave mirror, for example, is used in searchlights to direct the beam of light, in telescopes to gather more light from faraway objects, and in headlights to concentrate the beam of light to illuminate the road ahead.

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Answer:

1st is hydropower

2nd is fossil fuels

A man sees a magnified image of a closely kept object by holding a simple 'clear glass' device in his hand. this device is likely to be a convex lens .

Option C.

The device described in the scenario is a clear glass, which suggests that it is likely a lens rather than a mirror. Additionally, since the image is magnified, it is likely a converging lens that is being used to view the object.

Therefore, the device in question is most likely a convex lens (option c) that is being used as a simple magnifying glass.

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Answer:

The single touch method of making a magnet is a simple way to magnetize a ferromagnetic material, such as a piece of iron. The process involves taking a magnet and bringing it into contact with the material to be magnetized, while moving it along the length of the material in a single direction. This process aligns the magnetic domains within the material, causing it to become magnetized. The strength of the resulting magnet depends on the strength of the magnet used and the length of time it is in contact with the material.

The process of making an electromagnet involves using an electric current to create a magnetic field. To make an electromagnet, a wire is wound into a coil around a magnetic core, such as an iron rod. When an electric current flows through the wire, it creates a magnetic field around the coil, which in turn magnetizes the magnetic core. The strength of the resulting magnet depends on the strength of the current flowing through the wire, the number of coils in the wire, and the type of magnetic core used. Electromagnets are commonly used in a variety of applications, including electric motors, speakers, and MRI machines.

Explanation:

Answer:

[tex]a = - \frac{5}{4} \: m{s}^{ - 2}[/tex]

Explanation:

a = acceleration (m/s²)

v = final velocity (m/s) = 0

u = initial velocity (m/s) = 5

t = time (s) = 4

[tex]a = \frac{v - u}{t} \\ a = \frac{(0 - 5)}{4} \\ a = - \frac{5}{4} [/tex]

Answer:

Resistance to the flow of electricity in wires and resistance to the flow of water in pipes are similar in several ways:

Both are affected by the size of the conductor: In both cases, the resistance to flow is affected by the size of the conductor. A larger diameter wire or pipe will offer less resistance to the flow of electricity or water than a smaller diameter one.

Both are affected by the length of the conductor: The resistance to flow also increases with the length of the conductor. A longer wire or pipe will offer more resistance to the flow of electricity or water than a shorter one.

Both are affected by the material of the conductor: Different materials have different resistances to flow. In electrical wires, metals are commonly used as they have low resistance. In pipes, materials such as copper, plastic, or steel may be used depending on the specific application.

Both can be reduced by minimizing obstructions: Resistance to flow can be reduced in both cases by minimizing obstructions in the path of the flow. In pipes, for example, the use of smooth inner walls and proper pipe bends can help reduce resistance. In electrical wires, minimizing the number of connections and using high-quality connections can help reduce resistance.

Overall, resistance to flow is a fundamental concept that applies to the flow of electricity and water, and the factors affecting resistance are similar in both cases.

Answer:

On a wet marble floor, the friction is very less because marble floor has irregularities that get filled with water. We can walk only when there is more friction. Due to the less friction, we may slip. So, it is very difficult to walk on a wet marble floor.

Answer:

because marbles are always sliding and water makes them more slimy

When the same tuning fork is sounded together with the 266-hz tone, a beat frequency of 3 hz is produced, the frequency of the tuning fork is: 264 Hz.

When two out-of-tune flutes play the same note, and one produces a tone that has a frequency of 262 Hz, and the other produces 266 Hz. The beat frequency of 1 Hz is produced when a tuning fork is sounded together with the 262 Hz tone, and the same tuning fork is sounded together with the 266-Hz tone.

In this case, the frequency of the tuning fork is required. The formula for beat frequency is expressed as follows: Beat frequency = frequency 2 − frequency 1

The frequency of the tuning fork can be determined by subtracting the frequency of the 266 Hz tone from that of the 262 Hz tone (i.e. 266 - 262 = 4 Hz).

This implies that the tuning fork is 4 Hz higher than the 266 Hz tone. The beat frequency of 3 Hz when the same tuning fork is sounded together with the 266 Hz tone implies that the tuning fork's frequency is 3 Hz lower than the 269 Hz tone. The average of these two frequencies is 264 Hz (i.e. 266 Hz - 2 Hz). Therefore, the frequency of the tuning fork is 264 Hz.

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The  centripetal force the passenger experience is 30167.52 N

The centripetal force experienced by the passenger can be calculated using the formula:

Fc = (m)(r)(ω^2)

where Fc is the centripetal force, m is the mass of the passenger, r is the radius of the loop, and ω is the angular speed of the roller coaster.

Given:

m = 80 kg

r = 12 m

ω = 3.14 rad/s

Substituting these values into the formula, we get:

Fc = (80 kg)(12 m)(3.14 rad/s)^2

Fc = 30167.52 N

Therefore, the passenger experiences a centripetal force of approximately 30167.52 N when the roller coaster reaches an angular speed of 3.14 rad/s.

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The reason that the melting point of table salt is higher than that of sugar is A, the particles of salt are attracted more strongly than those of sugar.

Melting point is measured by heating a substance until it melts and then recording the temperature at which melting occurs. This is typically done using a device called a melting point apparatus or a melting point apparatus.

The sample is placed in a small capillary tube and inserted into the melting point apparatus, which heats the sample gradually and detects the temperature at which the first signs of melting are observed. The temperature is recorded as the melting point of the substance.

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Image transcribed and translated:

What is the reason that the melting point of table salt is higher than that of sugar?

A. The particles of salt are attracted more strongly than those of sugar.

B. The particles of sugar are attracted more strongly than those of salt.

C. The particles of salt are attracted more weakly than those of sugar.

D. The particles of sugar repel while those of salt attract.

The speed (in revolutions per minute) at which resonance vibrations will occur is 273.91 RPM.

The front-loading washing machine is mounted on a thick rubber pad that acts as a spring. When the rotor spins at radians per second, the rotor exerts a vertical force of newtons on the machine.

Resonance vibrations occur when the natural frequency of the machine matches the frequency of the external force applied to it. This can cause the machine to vibrate excessively and even damage it.

In this scenario, the natural frequency of the washing machine is related to its spring constant and the mass of the machine. Neglecting friction, resonance vibrations occur at RPMs.

Given,

The weight (with m/s) of the machine depresses the pad exactly 0.89 cm.

The rotor exerts a vertical force of N on the machine.

The natural frequency of a machine is given by:

f = 1/2π x √(k/m)

where f is the frequency of oscillation,

k is the spring constant of the spring,

m is the mass of the machine.

In this case, the weight of the machine depresses the spring by 0.89 cm.

So, the effective spring constant is given by,

k = F/x = (mg)/x

where F is the weight of the machine,

m is its mass,

g is the acceleration due to gravity,

and x is the amount by which the spring is compressed.

Substituting the given values,

k = (m x g) / x = (m x 9.8) / 0.0089 = 1101.12m N/m

Substituting the given values,

f = 1/2π x √(k/m) = 1/2π x √(1101.12m/m) = 4.9448 √m Hz

The frequency of oscillation is given in Hz.

To convert it to RPMs, we need to divide it by 60 (the number of seconds in a minute) and multiply it by 2π (the number of radians in a circle).

f = 4.9448 √m Hz = (4.9448 √m x 2π)/60 RPM

At resonance, the frequency of oscillation should match the frequency of the rotor.

So, the frequency of the rotor is given by,

f = ω/2π

where ω is the angular velocity of the rotor in radians per second.

Substituting the given values,

N = mg = F,

ω = N/(m x r) = 94.4912/r rads/s

where r is the radius of the rotor in meters.

Substituting these values in the equation for frequency,

f = ω/2π = (94.4912/r) / 2π = 15.0415 / r Hz

At resonance, the frequency of oscillation should match the frequency of the rotor.

Therefore,4.9448 √m = 15.0415 / r

Squaring both sides,19.4542 m = (15.0415/r)^2

Solving for r, r = 0.0275 m = 2.75 cm

Therefore, the radius of the rotor is 2.75 cm.

The speed (in revolutions per minute) at which resonance vibrations will occur is 273.91 RPM.

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If the water vapor comprises 0.2% of an air parcel and the pressure exerted by it is 1 mb, we can use the ideal gas law to determine the total atmospheric pressure of the parcel.

The ideal gas law states that:

PV = nRT

where P is the pressure of the gas, V is its volume, n is the number of moles of gas, R is the ideal gas constant, and T is the absolute temperature.

Assuming that the air in the parcel behaves as an ideal gas, we can use the ideal gas law to relate the pressure of the water vapor to the total atmospheric pressure of the parcel.

Let's assume that the total number of moles of gas in the parcel is N. Since the water vapor comprises 0.2% of the parcel, the number of moles of water vapor in the parcel is 0.002N. Therefore, the number of moles of dry air in the parcel is (1 - 0.002)N = 0.998N.

Using the ideal gas law, we can write:

(P_water vapor)(V) = (0.002N)(R)(T)

(P_dry air)(V) = (0.998N)(R)(T)

Adding these two equations together, we get:

P total can be solved as follows:

N(R)(T) / V Equals P total

(P_total)(V) = N(R)(T)

where P_total is the total atmospheric pressure of the parcel.

Therefore, we can solve for P_total as:

P_total = (N(R)(T)) / V

We can substitute the value of P_water vapor, which is 1 mb, into the first equation to find the volume of the water vapor in the parcel. Then, we can substitute the given values of N, R, and T, and the calculated volume of the water vapor into the equation for P_total to find the total atmospheric pressure of the parcel.

In summary, by using the ideal gas law and assuming that the air in the parcel behaves as an ideal gas, we can determine the total atmospheric pressure of the parcel given the pressure and volume of the water vapor, and the temperature and total number of moles of gas in the parcel.

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Even if the sun doubled in size, a person's weight on earth wouldn't alter.

If the sun somehow became twice as massive, your weight as normally measured here on earth would not change.

This is because weight is the force of gravity acting on an object, which depends on the mass of the object and the distance between the centers of mass of the two objects. While the mass of the sun has doubled, the distance between the centers of mass of the earth and the sun remains the same. Therefore, the gravitational force acting on the earth and on a person on the earth remains the same.

As a result, the weight of a person on the earth would not change if the sun became twice as massive.

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