An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.165×105 km and the planet has a radius of 4175 km and a mass of 6.70×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N·m2/kg2 .

Answers

Answer 1

Answer:

364days

Explanation:

Pls see attached file

Explanation:

An Undiscovered Planet, Many Light-years From Earth, Has One Moon, Which Has A Nearly Circular Periodic
Answer 2

The moon will take 112.7 days to make one revolution around the planet.

What is Kepler's third law?

The period of the satellite around any planet only depends upon the distance between the planet's center and satellite and also depends upon the planet's mass.

Given, the distance from the moon's center to the planet's surface,

h = 2.165 × 10⁵ km,

The radius of the planet, r = 4175 km  

The mass of the planet = 6.70 × 10²² kg

The total distance between the moon's center to the planet's center:

a = r +h = 2.165 × 10⁵ + 4175

a = 216500 + 4175

a = 220675

a = 2.26750 × 10⁸ m

The period of the planet can be calculated as:

[tex]T =2\pi \sqrt{\frac{a^3}{Gm} }[/tex]

[tex]T =2\3\times 3.14 \sqrt{\frac{(2.20675 \times 10^8)^3}{(6.67\times 10^{-11}).(6.70\times 10^{22})} }[/tex]

T = 9738253.26 s

T = 112.7 days

Learn more about Kepler's law, here:

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Related Questions

Calculate the range of wavelengths (in m) for AM radio given its frequency range is 540 to 1,600 kHz. smaller value m larger value m (b) Do the same for the ultraviolet frequency range of 760 to 30,000 THz. smaller value m larger value m

Answers

Answer:

a)  λ = 555.5 m,  λ = 187.5 nm

 

Explanation:

The velocity of a wave is given by the relation

           c = λ f

           λ = c /f

a) length of the radii AM

            λ = 3 10⁸/540 10⁻⁷

            λ= 5.555 102 m

            λ = 555.5 m

             f = 1600 kHz

            λ = 3 108/1600 103

             λ = 1.875 102 m

            lam  = 187.5 nm

b) light = 760 Thz = 760 10-12 Hz

               

         λ=  c/f

            λ = 3 108/760 10-12

            λ = 3.947 10-9

           λ= 30000 Thz

           λ= c/f

      λ = 3 10⁸/ 30000 10-12

       λ = 1  m

Grocery store managers contend that there is less total energy consumption in the summer if the store is kept at a low temperature. Make arguments to support or refute this claim, taking into account that there are numerous refrigerators and freezers in the store.

Answers

Answer:

Argument in favor of less total energy consumption if the store is kept at a low temperature

Explanation:

Have in mind that if the store has numerous refrigerators and freezers, the energy consumption of those machines have to be included into the analysis.

Recall that the efficiency (or Coefficient Of Performance - COP) of a frezzer or refrigerator is inversely proportional to the temperature difference between the inside of th machine and the environment where it is operation, therefore the smaller the difference, the highest their efficiency. Therefore, the cooler the environment (the temperature at which the store is kept) the better performance of the running refrigerators and freezers.

A cart rolls 2 m to the right then rolls back 1 m to the left.
a. What is the total distance rolled by the cart?

Answers

Explanation:

It is given that,

Distance covered by the cart to the right is 2 m

Distance covered by the cart to the left is 1 m

We need to find the total distance rolled by the cart. Total distance is equal to the sum of the distances covered by an object. It does depend on the direction.

Total distance = 2 m + 1 m

D = 3 m

The cart rolled to a total distance of 3 m.

The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various atmosphericprocesses, including lightning.
What is the excess charge on the surface of the earth? inC

Answers

Answer:

[tex]q = -461532.5 \ C[/tex]

Explanation:

From the question we are told that

     The  electric filed is  [tex]E = 102 \ N/C[/tex]  

Generally according to Gauss law

=>   [tex]E A = \frac{q}{\epsilon_o }[/tex]

Given that  the electric field is pointing downward  , the equation become

    [tex]- E A = \frac{q}{\epsilon_o }[/tex]

Here   [tex]q[/tex] is the excess charge on the surface of the earth

          [tex]A[/tex] is the surface  area of the of the earth which is mathematically represented as

     [tex]A = 4\pi r^2[/tex]

Where r is the radius of the earth which has a value [tex]r = 6.3781*10^6 m[/tex]

 substituting values

    [tex]A = 4 * 3.142 * (6.3781*10^6 \ m)^2[/tex]

    [tex]A =5.1128 *10^{14} \ m^2[/tex]

So

   [tex]q = -E * A * \epsilon _o[/tex]

Here [tex]\epsilon_o[/tex] s the permitivity of free space with value

          [tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

     [tex]q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}[/tex]

     [tex]q = -461532.5 \ C[/tex]

A rigid uniform bar of length L and mass m is suspended by a massless wire AC and a rigid massless link BC. Determine the tension in BC immediately after AC breaks.

Answers

Answer:

hello the needed diagram is missing attached below is the diagram and the detailed solution

The tension in BC = [tex]\frac{\sqrt{2} }{4} mg[/tex]

Explanation:

ATTACHED BELOW IS THE DETAILED SOLUTION T THE GIVEN PROBLEM

Ma = mg - T/ [tex]\sqrt{2}[/tex]  equation 1

Ma = 3T / [tex]\sqrt{2}[/tex]   equation 2

equate both equations to determine the tension on BC

CAN SOMEONE HELP ME PLEASE ITS INTEGRATED SCIENCE AND I AM STUCK

Answers

Answer:

[tex]\huge \boxed{\mathrm{Option \ D}}[/tex]

Explanation:

Two forces are acting on the object.

Subtracting 2 N from both forces.

2 N → Object ← 5 N

- 2 N                 - 2N

0 N → Object ← 3 N

The force 3 N is pushing the object to the left side.

The mass of the object is 10 kg.

Applying formula for acceleration (Newton’s Second Law of Motion).

a = F/m

a = 3/10

a = 0.3

on which principle does water pump work ?​

Answers

Answer:

The working principle of a water pump mainly depends upon the positive displacement principle as well as kinetic energy to push the water.

Explanation:

it mainly depends upon the positive displacement principle and also kinetic energy to push water. hope this hepls!

Describe how you expect the waveform and the sound you hear changes when you hit the tuning fork harder.

Answers

Answer:

In a tuning fork, two basic qualities of sound are considered, they are

1) The pitch of the waveform: This pitch depends on the frequency of the wave generated by hitting the tuning fork.

2) The loudness of the waveform: This loudness depends on the intensity of the wave generated by hitting the tuning fork.

Hitting the tuning fork harder will make it vibrate faster, increasing the number of vibrations per second. The number of vibration per second is proportional to the frequency, so hitting the tuning fork harder increase the frequency. From the explanation on the frequency above, we can say that by increasing the frequency the pitch of the tuning fork also increases.

Also, hitting the tuning fork harder also increases the intensity of the wave generated, since the fork now vibrates faster. This increases the loudness of the tuning fork.

Ibrah open a bottle of perfume infront of the room. After few minutes the smell of perfume reach the whole room. Explain why this happens​

Answers

the particles of the perfume began to spread into the air

3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing Photo 2.

Answers

Answer:

From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot  find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

[tex]E=\frac{F}{Q}[/tex]

But the force F

[tex]F= \frac{kQ1Q2}{r^2}[/tex]

But the electric field intensity due to a point charge Q at a distance r meters away is given by

[tex]E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }[/tex]

From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases

One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious space camper and you launch a serious rocket: it reaches an altitude of 211 km . What gain Δ???? in gravitational potential energy does the launch accomplish? The mass and radius of the Moon are 7.36×1022 kg and 1740 km, respectively.

Answers

Answer:

ΔP.E = 6.48 x 10⁸ J

Explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

ΔP.E = 6.48 x 10⁸ J

a person lifts 60kg on the surface of the earth, how much mass can he lift on the surface of the moon if he applies same magnitude of force​

Answers

Explanation:

Hey there!

According to the question;

A person can lift mass of 60 kg on earth.

mass(m1) = 60kg

acceleration due to gravity on earth (a) = 9.8m/s²

Now;

force (f) = m.a

= 60*9.8

= 588 N

Since, there is application of same magnitude of force on moon,

mass(m) =?

acceleration due to gravity on moon (a) = 1.67m/s²

Now;

force (f) = m.a

588 = m*1.67

m = 352.09 kg

Therefore, the person who can lift the mass of 60 kg on earth can lift mass of 352 kg on moon.

Hope it helps!

A flat loop of wire consisting of a single turn of cross-sectional area 7.30 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.50 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 2.60

Answers

Answer:

-0.73mA

Explanation:

Using amphere's Law

ε =−dΦB/ dt

=−(2.6T)·(7.30·10−4 m2)/ 1.00 s

=−1.9 mV

Using ohms law

ε=V =IR

I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA

Calculate the time it would take a cell phone signal to travel from a point on the equator to the satellite and back.

Answers

the signal to go back could possibly take 1 yr or maybe more

A mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 6 feet below the equilibrium position, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 1 2 the instantaneous velocity. Find the equation of motion x(t) if the mass is driven by an external force equal to f(t)

Answers

Answer:I don’t know

Explanation:

In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:_____.

Answers

Answer:

In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:  speed of light(c)

Explanation:

Generally the ratio of the E(electric field ) and  the B(magnetic field ) is  equal to the speed of the electromagnetic wave i.e the speed of  light (c) the value is

    [tex]c = 3.0 *10^{8} \ m/s[/tex]

A load of 25 kg is applied to the lower end and of a steal wire of length 25 m and thickness 3.0mm .The other end of wire is suspeded from a rigid support calculate strain and stress produced in the wire​

Answers

Answer:

the weight of the wire + 25kg

Explanation:

Why does cold water kept in an open container become warm on a hot summer afternoon?

Answers

Cold water kept in an open container acquires heat from the warmer surrounding becomes warm like the air around it due to the transfer of thermal energy

Pls mark as brainliest

Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the concentric spheres. When compared to the number of field lines N1 going through the sphere of radius R, the number of electric field lines N2 going through the sphere of radius 2R is

Answers

Answer:

N2 = ¼N1

Explanation:

First of all, let's define the terms;

N1 = number of electric field lines going through the sphere of radius R

N2 = number of electric field lines going through the sphere of radius 2R

Q = the charge enclosed at the centre of concentric spheres

ε_o = a constant known as "permittivity of the free space"

E1 = Electric field in the sphere of radius R.

E2 = Electric field in the sphere of radius 2R.

A1 = Area of sphere of radius R.

A2 = Area of sphere of radius 2R

Now, from Gauss's law, the electric flux through the sphere of radius R is given by;

Φ = Q/ε_o

We also know that;

Φ = EA

Thus;

E1 × A1 = Q/ε_o

E1 = Q/(ε_o × A1)

Where A1 = 4πR²

E1 = Q/(ε_o × 4πR²)

Similarly, for the sphere of radius 2R,we have;

E2 = Q/(ε_o × 4π(2R)²)

Factorizing out to get;

E2 = ¼Q/(ε_o × 4πR²)

Comparing E2 with E1, we arrive at;

E2 = ¼E1

Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;

N2 = ¼N1

A rope, under a tension of 153 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by . where at one end of the rope, is in meters, and is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation?

Answers

Complete question is;

A rope, under a tension of 153 N and fixed at both ends, oscillates in a second harmonic standing wave pattern. The displacement of the rope is given by

y = (0.15 m) sin[πx/3] sin[12π t].

where x = 0 at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c)the mass of the rope? (d) If the rope oscillates in a third - harmonic standing wave pattern, what will be the period of oscillation?

Answer:

A) Length of rope = 4 m

B) v = 24 m/s

C) m = 1.0625 kg

D) T = 0.11 s

Explanation:

We are given;

T = 153 N

y = (0.15 m) sin[πx/3] sin[12πt]

Comparing this displacement equation with general waveform equation, we have;

k = 2π/λ = π/2 rad/m

ω = 2πf = 12π rad/s

Since, 2π/λ = π/2

Thus,wavelength; λ = 4 m

Since, 2πf = 12π

Frequency;f = 6 Hz

A) We are told the rope oscillates in a second-harmonic standing wave pattern. So, we will use the equation;

λ = 2L/n

Since second harmonic, n = 2 and λ = L = 4 m

Length of rope = 4 m

B) speed is given by the equation;

v = fλ = 6 × 4

v = 24 m/s

C) To calculate the mass, we will use;

v = √T/μ

Where μ = mass(m)/4

Thus;

v = √(T/(m/4))

Making m the subject;

m = 4T/v²

m = (4 × 153)/24²

m = 1.0625 kg

D) Now, the rope oscillates in a third harmonic.

So n = 3.

Using the formula f = 1/T = nv/2L

T = 2L/nv

T = (2 × 4)/(3 × 24)

T = 0.11 s

A 70 kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above water when his lungs are full.

Required:
a. Calculate the volume of air he inhales - called his lung capacity - in liters.
b. Does this lung volume seem reasonable?

Answers

Answer:

Explanation:

A) Vair = 1.3 L

B) Volume is not reasonable

Explanation:

A)

Assume

m to be total mass of the man

mp be the mass of the man that pulled out of the water

m1 be the mass above the water with the empty lung

m2 be the mass above the water with full lung

wp be the weight that the buoyant force opposes as a result of the air.

Va be the volume of air inside man's lungs

Fb be the buoyant force due to the air in the lung

given;

m = 78.5 kg

m1 = 3.2% × 78.5 = 2.5 kg

m2 = 4.85% × 78.5 = 3.8kg

But, mp = m2- m1

mp = 3.8 - 2.5

mp = 1.3kg

So using

Archimedes principle, the relation for formula for buoyant force as;

Fb = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

Fb = wp = 1.3× 9.81

Fb = 12.7N

But

Fb = (ρ_water × V_air × g)

So

Vair = Fb/(ρ_water × × g)

Vair = 12.7/(1000 × 9.81)

V_air = 1.3 × 10^(-3) m³

convert to litres

1 m³ = 1000 L

Thus;

V_air = 1.3× 10^(-3) × 1000

V_air = 1.3 L

But since the average lung capacity of an adult human being is about 6-7litres of air.

Thus, the calculated lung volume is not reasonable

Explanation:

Which examination technique is the visualization of body parts in motion by projecting x-ray images on a luminous fluorescent screen?

Answers

Answer:

Fluoroscopy

Explanation:

A Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of an object. In its primary application of medical imaging, a fluoroscope allows a physician to see the internal structure and function of a patient, so that the pumping action of the heart or the motion of swallowing, for example, can be watched.

Please help!
Much appreciated!​

Answers

Answer:

your question answer is 22°

I’m pretty sure the answer is 22

What is the threshold velocity vthreshold(ethanol) for creating Cherenkov light from a charged particle as it travels through ethanol (which has an index of refraction of n

Answers

Explanation:

The velocity of light in a medium of refractive index [tex]n[tex] is given by,

[tex]v=\frac{c}{n}[/tex]

[tex]v \text { is the velocity of light in the medium }[/tex]

[tex]c \text { is speed of light in vacuum }[/tex]

The exact value of speed of light in vacuum is [tex]299792458 \mathrm{m} / \mathrm{s}[/tex].

For Cherenkov radiation to be emitted, the velocity of the charged particle traversing the medium must be greater than this velocity. Thus, the threshold velocity of for creating Cherenkov radiation is,

[tex]v_{\text {Cherenkov }} \geq \frac{c}{n}[/tex]

[tex]v_{\text {threshod }}=\frac{c}{n}[/tex]

For water [tex]n=1.33,[tex] thus the threshold velocity for producing Cherenkov radiation in water is,

[tex]v_{\text {threatold }(\text { water })} &=\frac{299792458 \mathrm{m} / \mathrm{s}}{1.33}[/tex]

[tex]=225407863 \mathrm{m} / \mathrm{s}[/tex]

[tex]=2.254 \times 10^{8} \mathrm{m} / \mathrm{s}[/tex]

For ethanol [tex]n=1.36[tex], thus the threshold velocity for producing Cherenkov radiation in water is,

[tex]v_{\text {threstold }( \text { ettanol) } } &=\frac{299792458 \mathrm{m} / \mathrm{s}}{1.36}[/tex]

[tex]=220435630 \mathrm{m} / \mathrm{s}[/tex]

[tex]=2.204 \times 10^{8} \mathrm{m} / \mathrm{s}[/tex]

Answer:

The answer is "2.2 × [tex]\bold{10^8}[/tex]".

Explanation:

In the given question the value of n is missing which can be defined as follows:

n= 1.36

The velocity value of the threshold(ethanol) for a generation the Cerenkov light from the charged particle by travel through ethanol as:

know we will have to use an equation as follows:

Formula:      

(ethanol) or the vthreshold = [tex]\frac{c}{n}[/tex]

                                         [tex]= \frac{3\times 10^8} {1.36} \\\\= 2.2 \times 10^8[/tex]

The water in vthreshold:

[tex]= 2.2 \times 10^8 \ \ \frac{m}{ s} \\\\[/tex]

Express the value in c, that is multiple, so, the value of vthreshold(water) is:

=(0.735) c

an alternating voltage of 100V, 50HZ Is Applied across an impedance of (20-j30) calculate the resistance, the capacitance, current, the phase angle between current and voltage

Answers

The resistance R = 20 Ω

The capacitance C = 106.1 μF

The current, I is 2.773 A at 56.31°.

The phase angle of the between the current and the voltage is 56.31° leading.

Since the impedance Z = 20 - j30 Ω, the resistance, R is the real part of the impedance. So R = ReZ = 20 Ω

So, the resistance R = 20 Ω

To find the capacitance, we need first to find the reactance of the capacitor X. Since the impedance Z = 20 - j30, the reactance of the capacitor X. is the imaginary part of the impedance. So X = ImZ = 30 Ω.

Now the reactance of the capacitor X = 1/ωC where ω = angular frequency of the circuit = 2πf where f = frequency of the circuit = 50 Hz and C = capacitance  

So, C = 1/ωX = 1/2πfX

Substituting the values of the variables into the equation, we have

C = 1/2πfX

C = 1/(2π × 50 Hz × 30 Ω)

C = 1/3000π

C = 1/9424.778

C = 1.061 × 10⁻⁴ F

C = 106.1 × 10⁻⁶ F

C = 106.1 μF

So, the capacitance is 106.1 μF

The current I = V/Z where V = voltage = 100 V at 0° and Z = impedance.

The magnitude of Z = √(20² + (-30)²)

= √(400 + 900)

= √1300

= 36.06 Ω

and its angle Φ = tan⁻¹(ImZ/ReZ)

= tan⁻¹(-30/20)

= tan⁻¹(-1.5) = -56.31°

So, V = 100 ∠ 0° and Z = 36.06 ∠ -56.31°

So, the current, I = V/Z =  (100 ∠ 0°)/36.06 ∠ -56.31°

= 100/36.06 ∠(0° - (-56.31° ))

= 2.773 ∠ 56.31° A

So, the current is 2.773 A at 56.31°.

Since the current is 2.773 A at 56.31°, the phase angle of the between the current and the voltage is 56.31° leading.

So, the phase angle of the between the current and the voltage is 56.31° leading.

Learn more about alternating voltage here:

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A balloon contains 1.21 x 105 L of ideal gas at 265K. The gas is then cooled to 201 K. What is the volume (L) assuming no gas enters or exits the balloon

Answers

Answer:

The new volume will be 0.918 x 10^5 L

Explanation:

initial volume  [tex]V_{1}[/tex] = 1.21 x 10^5 L

Initial temperature [tex]T_{1}[/tex] = 265 K

Final volume [tex]V_{2}[/tex] = ?

Final temperature [tex]T_{2}[/tex] = 201 K

Th gas is an ideal gas.

For ideal gases, the equation [tex]V_{1}[/tex]/[tex]T_{1}[/tex] = [tex]V_{2}[/tex]/[tex]T_{2}[/tex] = constant

substituting value, we have

(1.21 x 10^5)/265 = [tex]V_{2}[/tex]/201

[tex]V_{2}[/tex] =  24321000/265 = 91777.4 L

= 0.918 x 10^5 L

Several books are placed on a table. These books have a combined weight of 25 N and cover an area of 0.05 m2. How much pressure do the books exert on the table? The pressure the books apply to the table top is __ Pa.

Answers

Answer:

500 Pascals

Explanation:

[tex]Force = 25N\\Area = 0.05m^2\\\\Pressure = \frac{Force}{Area}\\ \\Pressure = \frac{25}{0.05}\\\\ Pressure = 500 Pascals[/tex]

For an object to move, a(n) _______ force must be applied. Question 1 options: Balanced Unbalanced

Answers

Answer:

Unbalenced

Explanation:

when balenced forces are applied to an object there is no motion. When you apply unbalenced force the object you are applying the force to will move in the opposite direction of the force.

Answer:

im pretty sure it unbalenced

Explanation:

i just am

A hydraulic lift raises a 2 000-kg automobile when a 500-N force is applied to the smaller piston. If the smaller piston has an area of 10 cm2, what is the cross-sectional area of the larger piston

Answers

Answer:

The cross-sectional area of the larger piston is 392 cm²

Explanation:

Given;

output mass of the piston, m₀ = 2000 kg

input force of the piston, F₁ = 500 N

input area of the piston, A₁ = 10 cm² = 0.001 m²

The output force is given by;

F₀ = m₀g

F₀ = 2000 x 9.8

F₀ = 19600 N

The cross-sectional area of the larger piston or output area of the piston will be calculated by applying the following equations;

[tex]\frac{F_i}{A_i} = \frac{F_o}{A_o} \\\\A_o= \frac{F_o A_i}{F_i} \\\\A_o = \frac{19600*0.001}{500} \\\\A_o = 0.0392 \ m^2\\\\A_o = 392 \ cm^2[/tex]

Therefore, the cross-sectional area of the larger piston is 392 cm²

Imagine two pairs of books. In the first pair, two books of mass m_1m 1 ​ m, start subscript, 1, end subscript and m_2m 2 ​ m, start subscript, 2, end subscript are separated by the distance rrr, resulting in a gravitational force F_1F 1 ​ F, start subscript, 1, end subscript. In the second pair, two books of mass 2m_12m 1 ​ 2, m, start subscript, 1, end subscript and 4m_24m 2 ​ 4, m, start subscript, 2, end subscript are separated by the distance 2r2r2, r, resulting in a gravitational force of F_2F 2 ​ F, start subscript, 2, end subscript. The relationship between these two forces can be written as F_2 = nF_1F 2 ​ =nF 1 ​ F, start subscript, 2, end subscript, equals, n, F, start subscript, 1, end subscript. In the expression F_2 = nF_1F 2 ​ =nF 1 ​ F, start subscript, 2, end subscript, equals, n, F, start subscript, 1, end subscript, what is the value of nnn?

Answers

The value of n in the relationship between F₁ and F₂, (F₂ = n·F₁), is n = 2

The reason for the above value is as follows;

The given parameters of the first pair of books are;

The masses of the books = m₁, and m₂

The distance between the two m₁ and m₂ = r

The gravitational force between the masses = F₁

The given parameters of the second pair of books are;

The masses of the books second pair = 2·m₁, and 4·m₂

The distance between the two masses in second pair = 2·r

The gravitational force between the masses in second pair = F₂

The relationship between the two forces is F₂ = n·F₁

The required parameter;

The value of n in F₂ = n·F₁

According to Newton's law of universal gravitation, we have;

[tex]\mathbf{F = G \times \dfrac{m_1 \times m_2}{r^2}}[/tex]

Therefore, we get;

[tex]F_1 = G \times \dfrac{m_1 \times m_2}{r^2}[/tex]

[tex]F_2 = G \times \dfrac{2\cdot m_1 \times 4 \cdot m_2}{(2 \cdot r)^2} = G \times \dfrac{8 \times m_1 \times m_2}{4 \times r^2} = 2 \times G \times \dfrac{ m_1 \times m_2}{r^2} = 2 \times F_1[/tex]

Therefore;

F₂ = 2·F₁

The value of n in F₂ = n·F₁ is n = 2

Learn more about gravitational force between masses here;

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