4. How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a space in between such that the cut edges are closest to each other

Answer:

The answer is option A.

Hope this helps you

Answer:

Answer is A Nuclear

Explanation:

Just answered this question on my test

Answer:

it attracts

Explanation:

since in a magnetic body there are two poles

(north and south poles)if the first pole was repeled when brought near the North Pole therefore the other end is going to attarct because the first end was also a North Pole while the second end will be a south pole

Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\[/tex]

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)][/tex]

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

[tex]R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C[/tex]

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

Answer:

The  angular speed is [tex]w = 5.89 \ rad/s[/tex]

Explanation:

From the question we are told that

    The time taken is  [tex]t = 1.6 s[/tex]

    The number of somersaults  is n  =  1.5

The total angular displacement during the somersault is mathematically represented as

         [tex]\theta = n * 2 * \pi[/tex]

substituting values

        [tex]\theta = 1.5 * 2 * 3.142[/tex]

       [tex]\theta = 9.426 \ rad[/tex]

 The angular speed is mathematically represented as

         [tex]w = \frac{\theta }{t}[/tex]

substituting values

         [tex]w = \frac{9.426}{1.6}[/tex]

          [tex]w = 5.89 \ rad/s[/tex]

Answer:

May be egg shaped

Has no new stars being formed.

Has almost no gas or dust between stars.

Explanation:

Elliptical galaxy is the collection of many stars which are bounded together gravitationally, which is smooth and ellipsoidal and shape and the appearance is featureless.

Elliptical galaxy is ovoid or spherical masses of stars.

It is found in galaxy clusters and compact galaxies.

It has no gas or dust between stars which result in low rates of star formation.

It is formed When two spirals collide, they lose their familiar shape, morphing into the less-structured elliptical galaxies.

Elliptical galaxy is made of old stars and have no gas and dust.

An example is elliptical galaxy m60 which shines brightly and is egg shaped.

Answer:

17.94 kN/C is the electric field intensity at the origin due to the charges.

Explanation:

From the question, we are told that

The distance of 1 μC from origin = 1 m

The distance of 2 μC from origin = 2 m

The distance of 3 μC from origin = 3 m

The distance of 4 μC from origin = 5 m

Therefore, for us to find the electric field intensity, we'll solve below:

The formula for Electric field intensity = ( k * q ) / ( r * r )

where , r is distance ,

k = 9 * 10^9 ,

and , q is charge .

now ,

electric field intensity at the origin = [ k * 10^(-6) / 1 * 1 ] +[ k * 2 * 10^(-6) / 2 * 2 ] + [ k * 3 * 10^(-6) / 3 * 3 ] + [ k * 4 * 10^(-6) / 5 * 5 ]

=> electric field intensity at the origin = k * 10^(-6) [ 1 + 1/2 + 1/3 + 4/25 ] N/C

=> electric field intensity at the origin = 9 * 10^9 * 10^(-6) * 1.99 N/C

=> electric field intensity at the origin = 17.94 kN/C

Answer:

The linear density is  [tex]K = 3.863 *10^{-3 } \ kg/m[/tex]

Explanation:

From the question we are told that

     The density of  steel is  [tex]\rho = 7800 \ kg/m^3[/tex]

      The diameter of the string is  [tex]d = 0.794 \ mm = 7.94 *10^{-4} \ m[/tex]

       The  radius of  the string is  evaluated as  [tex]r = \frac{D}{2} = \frac{7.94 *10^{-4}}{2} = 3.97*10^{-4} \ m[/tex]

The volume of the string is  mathematically evaluated as

       [tex]V = \pi * r ^2 * L[/tex]

Now assuming that the length of the string is  L = 2 m  

      So  

         [tex]V = 3.142 * (3.97 *10^{-4})^2 * (2)[/tex]

        [tex]V = 9.9041 *10^{-7} \ m^3[/tex]

Then the mass of the string would be  

       [tex]m = \rho * V[/tex]

substituting value  

       [tex]m = 7800*9.904 14 *10^{-7}[/tex]

      [tex]m = 7.73*10^{-3} \ kg[/tex]

Looking at the question we see that the unit of the linear density is  [tex]\frac{kg}{m}[/tex]

Hence the linear density is evaluated as

        [tex]K = \frac{m}{L}[/tex]

substituting value  

        [tex]K = \frac{7.73 *10^{-3}}{2}[/tex]

        [tex]K = 3.863 *10^{-3 } \ kg/m[/tex]

Answer:

The particle P moves directly upwards

Explanation:

Lets designate the negative point charge at point P as particle P

The +Q charge will exert an attractive force on the particle P.

The -Q charge will exert a repulsive force on the particle P

The +Q charge exerts an upwards and leftward force on particle P

The -Q charge exerts an upwards and rightward force on particle P

Since the charges are equidistant from the particle P, and are of equal magnitude, the rightward force and the leftward force will cancel out, leaving just the upward force on the particle P.

The effect of the upward force is that the particle P moves directly upwards

━━━━━━━☆☆━━━━━━━

▹ Answer

B. Anxiety

▹ Step-by-Step Explanation

Anxiety isn't a behavior since it's a feeling. Behavior and feeling are different things therefore, anxiety is the correct answer.

Hope this helps!

- CloutAnswers ❁

Brainliest is greatly appreciated!

━━━━━━━☆☆━━━━━━━

Eating, sleeping, and crying all are considered as behaviors. However, anxiety cannot be considered as a behavior because it is a feeling. Thus, the correct option is B.

Anxiety is an intense feeling of excessive, and persistent worry and the fear about everyday situations. This includes fast heart rate, rapid breathing, sweating, and feeling tired constantly may occur.

Behavior is the range of actions and mannerisms which are made by individuals, organisms, systems or the artificial entities in some environment. These systems can include other systems or organisms as well as the inanimate physical environment. Behaviors include eating, sleeping, and crying. Anxiety is not a behavior, it is a feeling.

Therefore, the correct option is B.

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Answer: The frequency heard is 562.7 Hz.

Explanation: Doppler Effect happens when there is shift in frequency during a realtive motion between a source and the observer of that source.

It can be calculated as:

[tex]f_{o} = f_{s}(\frac{c+v_{o}}{c-v_{s}} )[/tex]

where:

c is the speed of light (c = 332m/s)

all the subscripted s is related to the Source (frequency, velocity);

all the subscripted o is related to the Observer (frequency, velocity);

As the source is moving towards the observer and the observer is moving towards the source, the velocities of each are opposite related to direction.

So, the frequency perceived by the observer:

[tex]f_{o} = 439(\frac{332+32}{332-48} )[/tex]

[tex]f_{o} = 439(\frac{364}{284} )[/tex]

[tex]f_{o} = 439(1.282 )[/tex]

[tex]f_{o}[/tex] = 562.7 Hz

At this condition, the observer hears the train's horn in a perceived frequency of 562.7 Hz

Answer:

1000 m upwards

Explanation:

Displacement Formula: Average Velocity = Displacement/Total Time

Simply plug in our known variables and solve:

100 m/s = x m/10 seconds

100 m/s(10 s) = x m

m = 1000

The complete question is;

In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.

Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?

Answer:

F_top = 385.36 N

Explanation:

We are given;

mass;m = 52 kg

Time;t = 4.3 s

Diameter;d = 16m

So,Radius;r = 16/2 = 8m

The formula for the centrifugal force is given as;

F_c = mω²R

Where;

R = radius

Angular velocity;ω = 2πf

f = frequency = 1/t = 1/4.3 Hz

F_c = 53 × (2π × 1/4.3)² × 8 = 905.29 N.

The force at top would be;

F_top = F_c - mg

F_top = 905.29 - (9.81 × 53) N

F_top = 385.36 N

The force at the top of ride will be "385.36 N".

According to the question,

Rider's mass, m = 52 kg

Time, t = 4.3 s

Diameter, d = 16 m

Radius, r = [tex]\frac{16}{2}[/tex] = 8 m

Frequency, f = [tex]\frac{1}{t}[/tex] = [tex]\frac{1}{4.3}[/tex] Hz

We know the formula,

Centrifugal force,  [tex]F_c[/tex] = mω²R

or,

Angular velocity, ω = 2πf

By substituting the values in the above formula,

[tex]F_c = 53(2\pi \times (\frac{1}{4.3})^2\times 8 )[/tex]

    [tex]= 905.29[/tex] N

hence,

The top force will be:

→ [tex]F_{top} = F_c[/tex] - mg

By substituting the values,

          [tex]= 905.29-(9.81\times 53)[/tex]

          [tex]= 385.36[/tex] N

Thus the above response is correct.  

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Answer:

It would be 450 or 640. My final answer would be about 450

Explanation: Because it would't be to high if it was shot Voy = 100

btw i think i know what i know what i am talking about.

The answer would be about 450 m.

The top isn't the standard for a cliff to be reckoned as a cliff as such. Any steep rock face particularly at the edge of the sea can be specified as a cliff.

A 'clifftop' just refers to any pinnacle of a cliff. A 'plateau' is any flat extended geologic floor. An 'overhang' is a part of a structure or formation that protrudes from the primary frame and rests such that it is 'overhanging' the ground (striking above it).

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Explanation:

hope you like then comment plz

Answer:

The net charge on the shell is 30x10^-9C

Explanation:

Pls see attached file

Answer:

Option A. Li2O

Explanation:

To know which of the compound contains oppositely charged ions, let us determine the nature of each compound. This is illustrated below:

Li2O is an ionic compound as it contains a metal (Lithium, Li) and non metal (oxygen, O). Ionic compounds are charactized by the presence of aggregate positive and negative charge ions. This is true because they are formed by the transfer of electron(s) from the metallic atom to the non-metallic atom.

2Li —> 2Li^+ + 2e

O2 + 2e —> O^2-

2Li + O2 + 2e —> 2Li^+ + O^2- + 2e

2Li + O2 —> 2Li^+ O^2- —> Li2O

OF2 is a covalent compound as it contains non metals only (i.e oxygen, O and fluorine, F). Covalent compounds are characterised by the presence of molecules. This is true because they are formed from the sharing of electron(s) between the atoms involved.

PH3 is a covalent compound as it contains non metals only (i.e phosphorus, P and hydrogen, H).

SCl2 is a covalent compound as it contains non metals only (i.e sulphur, S and chlorine, Cl).

From the above information, we can see that only Li2O contains oppositely charged ions.

Answer:

A

Explanation:

Just took the test

Answer:

it is more reactive in high temperature than in low temperature.

Answer:

(C). The line integral of the magnetic field around a closed loop

Explanation:

Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux.

This can be written mathematically as;

[tex]EMF = -\frac{\delta \phi _B}{\delta t}[/tex]

[tex](\frac{\delta \phi _B}{\delta t} )[/tex] is the rate of change of the magnetic flux through a surface bounded by the loop.

ΔФ = BA

where;

ΔФ is change in flux

B is the magnetic field

A is the area of the loop

Thus, according to Faraday's law of electric generators

∫BdL = [tex]\frac{\delta \phi _B}{\delta t}[/tex] = EMF

Therefore, the line integral of the magnetic field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop.

The correct option is "C"

(C). The line integral of the magnetic field around a closed loop

Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal to: D. The flux of the electric field through a surface which has the loop as its boundary.

In Physics, the surface integral with respect to the normal component of a magnetic field over a surface is the magnetic flux through that surface and it is typically denoted by the symbol [tex]\phi[/tex].

Faraday's Law states that the negative of the time rate of change ([tex]\Delta t)[/tex] of the flux of the magnetic field ([tex]\phi[/tex]) through a surface is directly proportional to the flux ([tex]\phi[/tex]) of the electric field through a surface which has the loop as its boundary.

Mathematically, Faraday's Law is given by the formula:

[tex]E.m.f = -N\frac{\Delta \phi}{\Delta t}[/tex]

Where:

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Complete Question

The complete question is gotten from OpenStax

A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s ? The player’s mass is 70.0 kg, and air resistance is negligible.

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball

Answer:

The  time it will take is  [tex]t = 1.4907 \ s[/tex]  

Explanation:

From the  question we are told that

    The force experienced by the player is  [tex]F = 126 \ N[/tex]

     The distance of the ball from the player is  [tex]d = 2.00 \ m[/tex]

      The initial velocity is  u =  0 m/s because the player stopped

From the Newton law the acceleration of the player is mathematically evaluated as

             [tex]a = \frac{F}{m }[/tex]    [i,e  F =  ma  ]

substituting values

             [tex]a = \frac{126}{70}[/tex]

             [tex]a = 1.8 \ m/s^2[/tex]

Now from the equation of motion  we have that

           [tex]s = ut + \frac{1}{2} at^2[/tex]

substituting values              

             [tex]2.0 = 0 + \frac{1}{2} * 1.8 * t^2[/tex]

             [tex]t = \sqrt{ \frac{2.0}{0.9} }[/tex]

            [tex]t = 1.4907 \ s[/tex]

Answer:

The minimum speed required  is 2.62m/s

Explanation:

The value of  gravitational acceleration = g = 9.81 m/s^2

Radius of the vertical circle = R = 0.7 m

Given the mass of the pail of water = m

The speed at the highest point of the circle = V

The centripetal force will be needed must be more than the weight of the pail of water in order to not spill water.

Below is the calculation:

[tex]\frac{mV^{2}}{R} = mg[/tex]

[tex]V = \sqrt{gR}[/tex]

[tex]V = \sqrt{9.81 \times 0.7}[/tex]

[tex]V = 2.62 m/s[/tex]

Answer:

The lower frequency is [tex]f_1 = 265.55 \ Hz[/tex]

The higher frequency is  [tex]f_2 = 266.4546 \ Hz[/tex]

Explanation:

From the question we are told that

     The period is   [tex]T = 2.20 \ s[/tex]

      The frequency of the tuning fork is  [tex]f = 266.0 \ Hz[/tex]

Generally the beat frequency is mathematically represented as

       [tex]f_b = \frac{1}{T}[/tex]

substituting values

      [tex]f_b = \frac{1}{2.20}[/tex]

      [tex]f_b = 0.4546 \ Hz[/tex]

Since the beat  frequency is gotten from the beat produced by the tuning fork and and  the string   then

The possible frequency of the string ranges from

     [tex]f_1 = f- f _b[/tex]

to

    [tex]f_2 = f + f_b[/tex]

Now  substituting values

    [tex]f_1 = 266.0 - 0.4546[/tex]

    [tex]f_1 = 265.55 \ Hz[/tex]

For  [tex]f_2[/tex]

    [tex]f_2 = 266 + 0.4546[/tex]

    [tex]f_2 = 266.4546 \ Hz[/tex]

Answer:

The current is  [tex]I = 6.68 \ A[/tex]

Explanation:

From the question we are told that  

     The radius of the loop is  [tex]r = 6 \ cm = 0.06 \ m[/tex]

     The  earth's magnetic field is [tex]B_e = 0.7G= 0.7 G * \frac{1*10^{-4} T}{1 G} = 0.7 *10^{-4} T[/tex]

      The  number of turns is  [tex]N =1[/tex]

Generally the magnetic field generated by the current in the loop is mathematically represented as

        [tex]B = \frac{\mu_o * N * I}{2 r }[/tex]

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

         [tex]B = B_e[/tex]

=>     [tex]B_e = \frac{\mu_o * N * I }{ 2 * r}[/tex]

     Where  [tex]\mu[/tex] is the permeability of free space with value  [tex]\mu _o = 4\pi * 10^{-7} N/A^2[/tex]

       [tex]0.7 *10^{-4}= \frac{ 4\pi * 10^{-7} * 1 * I}{2 * 0.06}[/tex]

=>     [tex]I = \frac{2 * 0.06 * 0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}[/tex]

       [tex]I = 6.68 \ A[/tex]

The current in the loop will be "6.68 A".

According to the question,

Radius of loop, r = 6 cm or,

                           = 0.06 m

Earth's magnetic field, [tex]B_e[/tex] = 0.7 G or,

                                          = 0.7 × [tex]\frac{1\times 10^{-4}}{1 G}[/tex]

                                          = 0.7 × 10⁻⁴ T

Number of turns, N = 1

We know the relation,

→ B = [tex]\frac{\mu_0\times N\times I}{2r}[/tex]

or,

  B = [tex]B_e[/tex]

then,

→         [tex]B_e[/tex] = [tex]\frac{\mu_0\times N\times I}{2r}[/tex]

By substituting the values,

0.7 × 10⁻⁴ = [tex]\frac{4 \pi\times 10^{-7}\times 1\times I}{2\times 0.06}[/tex]  

hence,

The current will be:

               I = [tex]\frac{2\times 0.06\times 0.7\times 10^{-4}}{4 \pi\times 10^{-7}\times 1}[/tex]

                 = 6.68 A

Thus the above approach is correct.    

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Answer:

   M = 1433.5 kg

Explanation:

This exercise is solved using the Archimedean principle, which states that the hydrostatic thrust is equal to the weight of the desalinated liquid,

              B = ρ g V

with the weight of the truck it is in equilibrium with the push, we use Newton's equilibrium condition

           Σ F = 0

           B-W = 0

           B = W

       body weight

           W = M g

the volume is

           V = l to h

           rho_liquid g (l to h) = M g

           M = rho_liquid l a h

we calculate

            M = 1000 4.7 6.10 0.05

           M = 1433.5 kg

Answer:

363m.s-1

Explanation:

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\[/tex]

Now, determine the amplitude of oscillation, A;

[tex]E = \frac{1}{2} kA^2[/tex]

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

[tex]E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm[/tex]

Therefore, the amplitude of the oscillation is 2.82 cm

Answer:

0.0241 m

Explanation:

mass of the hockey player m1 = 90 kg

mass of puck m2 = 0.150 kg

puck velocity v1= 45 m/s

distance traveled by puck to reach the goal =15.0 m.

now accoding to momentum conservation law

90×45+0.15×v2 = 0 [ since, If both are initially at rest and if the ice is frictionless,]

therefore, v2= -0.0725 m/s.

Now time taken by the puck to reach the goal

t= 15/45 = 1/3 sec.

therefore, how far does the player recoil in the time

=0.0725×1/3= 0.0241 m.

the distance travelled by the player( recoil ) in the time the puck reach the goal is 0.025m.

Given the data in the question

Since they were both at rest initially

Using conservation of liner momentum:

[tex]mu + m_1u_1 = mv+ m_1v_1[/tex]

Now, Since they were both at rest initially

[tex]0 = mv+ m_1v_1[/tex]

We substitute in our values to find the velocity of the player after the hit ( recoil velocity )

[tex]0 =[ 0.150kg * 45.0m/s ] + [ 90.0kg * v_1 ]\\\\0 = 6.75kg.m/s + [ 90.0kg * v_1 ]\\\\90.0kg * v_1 = -6.75kg.m/s \\\\v_1 = -\frac{6.75kg.m/s}{90.0kg} \\\\v_1 =- 0.075m/s[/tex]

{ The negative sign shows that the velocity of both the player and the puck are in opposite direction }

Hence, recoil velocity of the player is 0.075m/s

Now, we determine the time taken for the puck to trach the goal using the relation between distance, velocity and time .

Time = Distance / Velocity

We substitute our values into the expression

[tex]t = \frac{s}{v} \\\\t = \frac{15.0m}{45m/s} \\\\t = 0.3333s[/tex]

Hence, the time taken for the puck to reach the goal is 0.3333 seconds.

Next, we determine the distance travelled by the player( recoil ) in the time the puck reach the goal using the relation between distance, velocity and time .

Time = Distance / Velocity

We substitute in our values

[tex]t = \frac{s}{v}\\\\0.3333s = \frac{s}{0.075m/s} \\\\s = 0.3333s * 0.075m/s\\\\s = 0.025m[/tex]

Therefore, the distance travelled by the player( recoil ) in the time the puck reach the goal is 0.025m.

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Answer:

Density = 10,933.93 kg/m^3

the density of the material is 10,933.93 kg/m^3

Explanation:

Density is the mass per unit volume

Density = mass/volume = m/V

Volume of a cylinder V = πr^2 h

Given;

Height h = 48.5mm = 0.0485 m

Radius r = diameter/2 = 49mm÷2 = 24.5mm = 0.0245m

Substituting the values;

Volume V = π×(0.0245^2)×0.0485

V = 0.000091458438030 m^3

V = 0.000091458 m^3

The mass is given as;

Mass = 1 kg

So, the density can be calculated as;

Density = 1/0.000091458

Density = 10933.92825785 kg/m^3

Density = 10,933.93 kg/m^3

the density of the material is 10,933.93 kg/m^3

Answer:

α(0) = 0 rad/s²

α(5) = 15 rad/s²

Explanation:

The angular velocity of the flywheel is given as follows:

w(t) = A + B t²

where, A and B are constants.

Now, for the angular acceleration, we must take derivative of angular velocity with respect to time:

Angular Acceleration = α (t) = dw/dt

α(t) = (d/dt)(A + B t²)

α(t) = 2 B t

where,

B = 1.5

AT t = 0 s

α(0) = 2(1.5)(0)

α(0) = 0 rad/s²

AT t = 5 s

α(5) = 2(1.5)(5)

α(5) = 15 rad/s²

Answer:

The ratio of the new force over the original force is 16

Explanation:

Recall the formula for the gravitational force between two masses M1 and M2 separated a distance D:

[tex]F_G=G\,\frac{M_1\,\,M_2}{D^2}[/tex]

So now, if the masses M1 and M2 are quadrupled and the distance stays the same, the new force becomes:

[tex]F'_G=G\,\frac{4M_1\,\,4M_2}{D^2}=G\,\frac{16\,\,M_1\,\,M_2}{D^2}=16\,\,G\,\frac{M_1\,\,M_2}{D^2}= 16\,\,F_G[/tex]

which is 16 times the original force.

So the ratio of the new force over the original force is 16

The ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance between them stays the same is 16:1.

Newton's law of gravitation states that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.

The formula for Newton's law of gravitation is:

[tex]F = G \frac{m_1m_2}{r^{2} }[/tex]

where,

The initial force between the planets is:

[tex]F_1 = G \frac{m_1m_2}{r^{2} }[/tex]

The force between the planets if the masses of both planets are quadrupled but the distance between them stays the same is:

[tex]F_2 = G \frac{4m_14m_2}{r^{2} } = 16 G \frac{m_1m_2}{r^{2} }[/tex]

The ratio of F₂ to F₁ is:

[tex]\frac{F_2}{F_1} =\frac{16 G \frac{m_1m_2}{r^{2} }}{G \frac{m_1m_2}{r^{2} }} = \frac{16}{1}[/tex]

The ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance between them stays the same is 16:1.

Learn more about Newton's gravitational law here: https://brainly.com/question/9373839