As John throws the ball 3 meters and the cat collects the ball and returns to John. Then. the distance traveled by the cat when it returns from the second time is calculated as
[tex]\begin{gathered} d=3+3 \\ =6\text{ m} \end{gathered}[/tex]3. The mass of an electron is 9.1x10^-31 kg. The mass of a proton is 1.7x10^-27 kg.The gravitational force between them in the hydrogen atom is 1.0x10^-47 N.What is the separation distance between them?
Given:
The mass of the electron is,
[tex]m_e=9.1\times10^{-31}\text{ kg}[/tex]The mass of the proton is,
[tex]m_p=1.7\times10^{-27}\text{ kg}[/tex]The gravitational force between the electron and the proton in the hydrogen atom is,
[tex]1.0\times10^{-47}\text{ N}[/tex]To find:
the separation between them
Explanation:
The gravitational force between two masses is,
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]Here, the universal gravitational constant is,
[tex]G=6.67\times10^{-11}\text{ Nm}^2kg^{-2}[/tex]Substituting the values we get,
[tex]\begin{gathered} 1.0\times10^{-47}=\frac{6.67\times10^{-11}\times9.1\times10^{-31}\times1.7\times10^{-27}}{r^2} \\ r^2=\frac{6.67\times10^{-11}\times9.1\times10^{-31}\times1.7\times10^{-27}}{1.0\times10^{-47}} \\ r^2=1.03\times10^{-20} \\ r=\sqrt{1.01\times10^{-20}} \\ r=1.01\times10^{-5}\text{ m} \end{gathered}[/tex]A speedboat increases its speed from 13.5 m/s to 27.4 m/s in a distance of 241 m. Determine the time over which this acceleration occurs?
V² = U² + 2aS
27.4² = 13.5² + 2*241a
750.76 = 182.25 + 482a
750.76 - 182.25 = 482a
568.51/482
a = 1.18m/s^2
A hair dryer draws 6.0 A when plugged into a 120-V line. (a) What is its resistance? (b) How much charge passes through it in 15 min?
We will have the following:
a)
[tex]R=\frac{120V}{6A}\Rightarrow R=20\Omega[/tex]b)
[tex]Q=(6A)(15min\ast\frac{60s}{1min})\Rightarrow Q=5400W[/tex]So, the charge is 5400 W in 15 minutes.
A mass falls from rest to the growl in 15 secs. Find y, the vertical distance it travelled
We will determine the distance fallen as follows:
[tex]d=v_it+\frac{1}{2}at^2[/tex]Thus:
[tex]d=(m/s)(15s)+\frac{1}{2}(9.8m/s^2)(15s)^2\Rightarrow d=1102.5m[/tex]So, it traveled 1102.5 meters.
it is known that the mass of the earth is 81 times the mass of the moon. show that the point of weightlessness between the earth and the moon for a spacecraft house occurs at √9/10 of the distance to the moon
Q4. A 5.0-kg bowling ball rolls down africtionless 2.5 m tall ramp and strikes astationary mass at the bottom of the ramp in aperfectly elastic collision. To whatheight back up the ramp does the first bowlingtravel after the collision ita) the stationary mass is also 5.0 kgb) the stationary mass is 10.0 kgC) the stationary mass is 500.00 kg
a)
Using conservation of energy for ball 1:
[tex]\begin{gathered} E1=E2 \\ K1+U1=K2+U2 \\ 0+m1gh=\frac{1}{2}m1v1^2 \end{gathered}[/tex]Solve for v1:
[tex]\begin{gathered} v1=\sqrt[]{2gh} \\ v1=\sqrt[]{2(9.8)(2.5)} \\ v1=7\frac{m}{s} \end{gathered}[/tex][tex]v1=v2^{\prime}-v1^{\prime}[/tex]Using conservation of momentum:
[tex]\begin{gathered} m1v1=m1(v2^{\prime}-v1^{\prime})+m2v2^{\prime} \\ v2^{\prime}=\frac{2m1v1}{(m1+m2)} \\ \end{gathered}[/tex]a)
m2 = 5 kg
[tex]v2^{\prime}=7\frac{m}{s}[/tex]so:
[tex]\begin{gathered} v1^{\prime}=7-7 \\ v1^{\prime}=0 \end{gathered}[/tex]so:
[tex]\begin{gathered} m1gh^{\prime}=\frac{1}{2}m1(v1^{\prime})^2 \\ h^{\prime}=0 \end{gathered}[/tex]If the stationary mass is 5.0 kg the height back up the ramp is 0 meters
b)
m2 = 10
[tex]v2^{\prime}=\frac{14}{3}\frac{m}{s}[/tex]so:
[tex]v1^{\prime}=-\frac{7}{3}\frac{m}{s}[/tex][tex]\begin{gathered} m1gh^{\prime}=\frac{1}{2}m1(v1^{\prime})^2 \\ h^{\prime}\approx0.2778m \end{gathered}[/tex]If the stationary mass is 10.0 kg the height back up the ramp is 0.2778 meters
c)
m2 = 500.00kg
[tex]v2^{\prime}=\frac{14}{101}\frac{m}{s}[/tex]so:
[tex]v1^{\prime}\approx-\frac{693}{101}\frac{m}{s}[/tex][tex]\begin{gathered} m1gh^{\prime}=\frac{1}{2}m1(v1^{\prime})^2 \\ h^{\prime}\approx2.4m \end{gathered}[/tex]If the stationary mass is 500.0 kg the height back up the ramp is 2.4 meters
McKenna is performing an experiment. She is testing the relationship between the temperature of water, and how long it takes for rust to form on iron plates. If she finds that higher temperatures lead to lower times for rusting, what is the qualitative relationship?positive correlationno correlationnegative correlationinverse square correlation
As with the increase in the temperature, the time taken to form the rust on the iron plate decreases.
This correlation is known as a negative correlation.
Thus, the relationship between the temperature of the water and the time taken to form rust on the iron plate is a negative correlation.
Hence, the third option is the correct answer.
A block of mass of 1kg is hit by a force of magnitude 5 N which causes the block to have an acceleration of magnitude a. If the same block is hit by a force of magnitude 40 N, then which of these is the new magnitude of acceleration of the block?A) 2.8284aB) one eighth of aC) one/ 64 of aD) 8aE) 64a
The acceleration is related to the force and mass by Newton's second law which states:
[tex]a=\frac{F}{m}[/tex]for the first mass we have:
[tex]a_1=\frac{5}{1}=5[/tex]If the same block is hit by a force of 40 N then the new acceleration is:
[tex]a=\frac{40}{1}=40[/tex]Which is eight times the original acceleration; therefore, the answer is D.
If a bullet leaves the muzzle of a rifle with a speed of 600 m/s, and the barrel of the rifle is 0.800 m long, at what rate is the bullet accelerated while in the barrel?That is the question I am having trouble with
Given that the final velocity of the bullet is
[tex]V_f=600\text{ m/s}[/tex]The initial velocity of the bullet is
[tex]V_i=0\text{ m/s}[/tex]The distance traveled by the bullet is x = 0.8 m
We have to find the acceleration.
Let the acceleration be denoted by a.
The equation which can be used to calculate acceleration is
[tex](V_f)^2-(V_i)^2=2ax[/tex]Substituting the values, acceleration will be
[tex]\begin{gathered} (600)^2-0^2=2\times a\times0.8 \\ a=\frac{360000}{1.6} \\ =225000m/s^2 \end{gathered}[/tex]Thus the acceleration will be 225000 m/s^2.
A uniform disk of radius 0.473 m and unknown mass is constrained to rotate about a perpendicular axis through its center. A ring with the same mass as the disk is attached around the disk's rim. A tangential force of 0.203 N applied at the rim causes an angular acceleration of 0.125 rad/s2. Find the mass of the disk.
From the calculations, the mass of the object is 3.43 Kg.
What is the mass of the rod?We know that the term circular motion has to do with motion along a circular path. The force that acts on the body in this case is tangential to the circle. This tangential force can be obtained from;
F = mrα
F = tangential force
m = mass of the object
r = radius of the object
α = angular acceleration
When we make m the subject of the formula;
m = F/rα
m = 0.203 N/0.473 m * 0.125 rad/s2
m = 3.43 Kg
Learn more about tangential force:https://brainly.com/question/13885495
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If raindrops are falling vertically at 7.90 m/s, what angle from the vertical do they make for a person jogging at 2.47 m/s? (Enter your answer in degrees.)
17.36 degrees
Explanation
Step 1
Diagram:
so, we have a rigth triangle then let
[tex]\begin{gathered} angle\text{ =}\theta \\ opposite\text{ side= 2.47} \\ adjacent\text{ side =7.9} \end{gathered}[/tex]Step 2
now, we need a function that relates those values to find the missin angle , it is
[tex]tan\theta=\frac{opposite\text{ side}}{adjacent\text{ side}}[/tex]replace ans solve for the angle
[tex]\begin{gathered} tan\theta=\frac{2.47}{7.9} \\ tan\theta=0.31 \\ inverse\text{ tan function in both sides} \\ \tan^{-1}(tan\theta)=\tan^{-1}(0.31) \\ \theta=17.36\text{ \degree} \end{gathered}[/tex]therefore, the answer is
17.36 degrees
I hope this helps you
Can you please answer the questions and explain how you got it?
Given data:
The mass of the block on table is M=500 g.
The hanging mass is m=300 g.
The kinetic friction is μ=0.25.
Part (a)
The free body diagram of the block on table is shown below,
The free body diagram of hanging mass is shown below,
Part (b)
The equation according Newton's second law for 500 g mass will be,
[tex]\begin{gathered} F_{net}=Ma \\ T-f=Ma \\ T-\mu N=Ma \\ T-\mu Mg=Ma \\ T=M(a+\mu g)\ldots\ldots(1) \end{gathered}[/tex]Here, f is the friction force, N is the normal force, and T is the tension in the string.
Part (c)
The equation according Newton's second law for 300 g mass will be,
[tex]\begin{gathered} F_{net}=ma \\ mg-T=ma \\ T=mg-ma \\ T=m(g-a)\ldots\ldots(2) \end{gathered}[/tex]Part (d)
Equating equation (1) and (2) to solve for acceleration,
[tex]\begin{gathered} M(a+\mu g)=m(g-a) \\ 500(a+0.25(9.81))=300(9.81-a) \\ a+2.45=0.6(9.81-a) \\ a+2.45=5.886-0.6a \\ a+0.6a=5.886-2.45 \\ 1.6a=3.436 \\ a=2.14m/s^2 \end{gathered}[/tex]Thus, the acceleration of the blocks are 2.14 m/s².
Part (e)
Substitute the value of acceleration in equation (1) to calculate the tension,
[tex]\begin{gathered} T=M(a+\mu g) \\ T=500(2.14+(0.25)(9.81)) \\ T=2296gm/s^2\times\frac{1\text{ N}}{1000gm/s^2} \\ T=2.296\text{ N} \end{gathered}[/tex]Thus, the tension in the sting is 2.296 N.
Hey there, I have a physics question that sadly I can't figure out since the pearson e book keeps crashing. Also I am blind and CAN'T SEE PICTURES OR GRAPHS!! So for the question: Let θ be the angle that the vector A⃗ makes with the +x-axis, measured counterclockwise from that axis. Find the angle θ for a vector that has the following components.Part AAx= 4.20 m, Ay= -2.10 mExpress your answer in degrees.
ANSWER:
333.4°
STEP-BY-STEP EXPLANATION:
To find angle for a vector:
[tex]\theta=\tan ^{-1}\mleft(\frac{A_y}{A_x}\mright)[/tex]We substitute the values of this case and the angle would then be:
[tex]\begin{gathered} \theta=\tan ^{-1}\mleft(\frac{-2.10}{4.20}\mright) \\ \theta=\tan ^{-1}(-0.5) \\ \theta=-26.56\cong-26.6 \\ \theta=360-26.6 \\ \theta=333.4\text{\degree} \end{gathered}[/tex]The angle is 333.4°
How do I do this problem? 9.A) A 100 g apple is falling from a tree. What is the impulse that Earth exerts on it during the first 0.5s ofits fall? What about the next 0.5 s?9.B) The same 100 g apple is falling from the tree. What is the impulse that Earth exerts on it in the first0.5 m of its fall? What about the second 0.5 m?9.c) Give a clear explanation for why the answers from 9.a and 9.b are different.
Given:
The mass of the apple is m = 100g = 0.1 kg
To find (A) the impulse during the first 0.5 s and in the next 0.5 s
(B) Impulse during the first 0.5 m of its fall and about the second 0.5 m.
(C)
Explanation:
(A) The force acting on the apple will be
[tex]\begin{gathered} F=mg \\ =0.1\times9.8\text{ } \\ =\text{ 0.98 N} \end{gathered}[/tex]Impulse during the first 0.5 s will be
[tex]\begin{gathered} Impulse\text{ = 0.98}\times0.5 \\ =0.49\text{ N s} \end{gathered}[/tex]Impulse during the second 0.5 s will be
[tex]\begin{gathered} Impulse\text{ =0.98}\times(0.5+0.5) \\ =0.98\text{ N s} \end{gathered}[/tex](B) The distance traveled by the apple is d = 0.5 m
[tex]\begin{gathered} d1=\frac{1}{2}g(t1)^2 \\ t1=\sqrt{\frac{2d1}{g}} \\ =\sqrt{\frac{2\times0.5}{9.8}} \\ =0.319\text{ s} \end{gathered}[/tex]The velocity will be
[tex]\begin{gathered} v1=gt1 \\ =9.8\times0.319 \\ =3.1262\text{ m/s} \end{gathered}[/tex]The distance traveled by the apple in the second 0.5 m
[tex]\begin{gathered} d2=\frac{1}{2}g(t2)^2 \\ t2=\sqrt{\frac{2d2}{g}} \\ =\sqrt{\frac{2\times0.5}{9.8}} \\ =0.319\text{ s} \end{gathered}[/tex]The velocity will be
'
[tex]\begin{gathered} v2=v1+gt2 \\ =3.1262+(9.8\times0.319) \\ =6.2524\text{ m/s} \end{gathered}[/tex]The impulse will be
[tex]\begin{gathered} Impulse=\text{ change in momentum} \\ =mv2-mv1 \\ =0.1\times(6.2524)-0.1\times(3.1262) \\ =0.62524-0.31262 \\ =0.31262\text{ N s} \end{gathered}[/tex](C) Although the numerical value is the same in both the cases but in part A it is the time and in part B it is the distance.
Layla was hiking in the Grand Canyon with her best friend Saige. The temperature when they started gut was only 4.0° C. She shouted at the canyon wall and 2.8 s later she heard her own voice echoing back. How far away was the canyon wall? a) 4.6 x 10² m b) 46.6 m c) 47 m d) 466.7 m
Given:
The temperature of the surrounding is 4 degrees Celsius.
The time taken for the echo to reach the ears is t = 2.8 s
Required:
The distance from the canyon wall.
Explanation:
The distance traveled by the sound is 2d as it travels up to the wall and bounces back.
The distance can be calculated by the formula
[tex]d=\frac{v\times t}{2}[/tex]Here, the speed of sound at 4 degrees Celsius is
[tex]v=331.3\text{ m/s}[/tex]On substituting the values, the distance traveled will be
[tex]\begin{gathered} d=\frac{331.3\times2.8}{2} \\ =4.6\times10^2\text{ m} \end{gathered}[/tex]Final Answer: The distance traveled is 4.6 x 10^2 m
- (True/False) If there is no friction, the efficiency of a machine can be greater than 100%.
ANSWER:
False
STEP-BY-STEP EXPLANATION:
Even without friction, the efficiency cannot be greater than 100%, because the output cannot be greater than the input, therefore, the statement is false.
Which of the following representsthis number in standard notation?5.05 · 104A. 50500B. 5050C. 0.000505D. 0.0505
The number is denoted in standard notation is as follows:
[tex]\begin{gathered} Y=5.05\times10^4 \\ Y=5.05\times10000 \\ Y=50500 \end{gathered}[/tex]Thus, the standard notation of this number is 50500.
Thus, the correct option for the above question is A.
A wave traveling on a Slinky® that is stretched to 4 m takes 4.97 s to travel the length of the Slinky and back again.(a) What is the speed (in m/s) of the wave? 1.61 m/s b) Using the same Slinky® stretched to the same length, a standing wave is created which consists of seven antinodes and eight nodes. At what frequency (in Hz) must the Slinky be oscillating? Hz =
Given:
The length of the slinky is: L = 4 m.
The time taken by the wave to travel the length and back again is: t = 4.97 s
To find:
a) The speed of the wave
b) The frequency of the wave
Explanation:
a)
As the wave on the slinky travels along the length and back again, it covers a distance that is double the distance of the slinky.
Thus, the total distance "d" traveled by the wave will be 2L.
The speed "v" of the wave is given as:
[tex]\begin{gathered} v=\frac{d}{t} \\ \\ v=\frac{2L}{t} \end{gathered}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} v=\frac{2\times4\text{ m}}{4.97\text{ s}} \\ \\ v=\frac{8\text{ m}}{4.97\text{ s}} \\ \\ v=1.61\text{ m/s} \end{gathered}[/tex]Thus, the speed of the wave is 1.61 m/s
b)
The standing wave created consists of seven antinodes and eight nodes. Thus, the length of the slinky is 7/2 times the wavelength of the wave.
[tex]L=\frac{7}{2}\lambda[/tex]Rearranging the above equation, we get:
[tex]\lambda=\frac{2}{7}L[/tex]Substituting the values in the above equation, we get:
[tex]\lambda=\frac{2}{7}\times4\text{ m}=\frac{8\text{ m}}{7}=1.143\text{ m}[/tex]The speed "v" of the wave is related to its wavelength "λ" and a frequency "f" as:
[tex]v=f\lambda[/tex]Rearranging the above equation, we get:
[tex]f=\frac{v}{\lambda}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} f=\frac{1.61\text{ m/s}}{1.143\text{ m}} \\ \\ f=1.41\text{ Hz} \end{gathered}[/tex]Thus, the frequency of the wave on the slinky is 1.41 Hz.
Final answer:
a) The speed of the wave is 1.61 m/s.
b) The frequency of the oscillation of the slinky is 1.41 Hz.
wut is phisics\and what do the do in physica
Physics is the branch of science that deals with the energy and matter and how they are relate to each other. Physics is simply as a natural science that depends on the approximation, measurements and mathematical analysis.
In this example, the system gains the ability to cause change the higher the brick is raised above the earth. Is this true or false?
ANSWER
True
EXPLANATION
The higher the brick is, the more potential energy it has. Therefore, it is true that it gains the ability to cause change the higher it is raised above the Earth.
Assume that the ammeter in the figure below is removed and the current that flows through the 4.0Ω path, I3, is unknown. Determine all the currents in the circuit.
The circuit is in parallel connection
Equivalent resistance = 1/Req = 1/R1 + 1/R2 + 1/R3
From the information given,
R1 = 5
R2 = 2
R3 = 4
1/Req = 1/5 + 1/2 + 1/4 = (4 + 10 + 5)/20 = 19/20
Req = 20/19 = 1.053 ohms
I = V/R
Given that V = 12,
Current flow through circuit = 12/1.053 = 11.4 A
I1 + I2 + I3 = 11.4
I1 = 12/5 = 2.4 A
I2 = 12/2 = 6 A
I3 = 12/4 = 3A
Define electricity in detail.
Electricity is the phenomenon of the flow of electric charges due to the potential difference.
The flow of electric charges across the conductor is known as electric current.
Unit of electric current is Ampere.
Potential difference is the difference of potential between two charges.
There are two charges namely positive and negative.
Uneven distribution of charges causes the potential difference.
Suppose A and B are two points on a conductor where A has a 5 C charge and B has a 3 C charge.
Thus, the current will flow from higher potential to lower potential which is from A to B.
Resistance is the opposing property of the conductor which opposes the flow of current.
The unit of resistance is ohm.
The frequency of a sound wave changes if the source of the sound is moving relative to the listener. What is this called?
Given
The frequency of a sound wave changes if the source of the sound is moving relative to the listener.
To find
What is this called?
Explanation
For an observer, when any source of sound is moving towards it or away from it then the sound becomes higher and lower respectively.
Thus the moment of the source changes the frequency of the sound. This is known as doppler effect.
Conclusion
The given effect is known as Doppler effect.
Charge q1 = 2.00 x 10-9 C is located +0.020 m from the origin along the x-axis. Charge q2 =-3.00 x 10-9 C is located +0.040 m from the origin along the x-axis. What is the electric force exerted by these two charges on a third charge, q3 = 5.00 x 10-9 C, located at the origin?
Given,
Charge q1 = 2.00 x 10-9 C is located +0.020 m from the origin along the x-axis. Charge q2 =-3.00 x 10-9 C is located +0.040 m from the origin along the x-axis.
There is a third charge q3 = 5.00 x 10-9 C, located at the origin.
Now the force acting on the third charge due to other two charge is:
[tex]\begin{gathered} F=k\frac{2\times10^{-9}\times5\times10^{-9}}{0.02^2}+k\frac{3\times10^{-9}\times5\times10^{-9}}{0.04^2} \\ \Rightarrow F=0.0002225+0.0000834=0.000305C \end{gathered}[/tex]
y=a on x Represents_____________
Answer:
a
Explanation:
to be linear equation, x needs to be multiplied by a constant but xy = a doesn't satisfy that
The binding energy of a nucleus is always negative.Question 4 options:TrueFalse
ANSWER
False.
EXPLANATION
Nuclear binding energy is the energy that is required to split the nucleus (of an atom) into nucleons: protons and neutrons.
The binding energy of a nucleus is always positive because nuclei require energy to separate them and it is impossible for nuclei to gain energy by being separated.
Therefore, the answer is false.
The correct unit for the frequency of oscillation is asecond/wavewave/secondHertzsecond
To find the unit of frequency.
Explanation:
The SI unit of frequency is Hertz (Hz). It is named after Heinrich Hertz.
It is equivalent to
[tex]\begin{gathered} frequency\text{ =}\frac{1}{time\text{ period}} \\ 1\text{ Hz}=\frac{1}{s} \\ =s^{-1} \end{gathered}[/tex]
A full tea cup has a mass of 0.40 kg. If the full cup applies a pressure of 1000.the radius of the circular ring imnrinted on the table?
Given:
The mass of the cup, m=0.40 kg
The pressure applied by the cup, P=1000 N/m²= 1000 Pa
To find:
The radius of the ring imprinted on the table.
Explanation:
The pressure is defined as the force per unit area.
Thus the pressure applied by the cup is given by,
[tex]\begin{gathered} P=\frac{F}{A} \\ =\frac{mg}{\pi r^2} \end{gathered}[/tex]Where A is the area of the ring, g is the acceleration due to gravity, and r is the radius of the ring.
On rearranging the above equation,
[tex]r=\sqrt{\frac{mg}{P\pi}}[/tex]On substituting the known values,
[tex]\begin{gathered} r=\sqrt{\frac{0.40\times10}{1000\pi}} \\ =0.036\text{ m} \\ =3.6\text{ cm} \end{gathered}[/tex]Final answer:
Thus the radius of the ring imprinted on the table is 3.6 cm
Therefore the correct answer is option 3.
A jet is goes from 180km/ to 139km/s in. 22 seconds what is its acceleration Vi-Vf-T-A-
Given
Initial velocity,
[tex]v_i=180\text{ km/s}[/tex]Final velocity,
[tex]v_f=139\text{ km/s}[/tex][tex]Time\text{ taken , t=22 s}[/tex]To find
The acceleration
Explanation
Acceleration is the ratio of change in velocity to that of the time taken
Thus,
[tex]\begin{gathered} a=\frac{v_f-v_i}{t} \\ \Rightarrow a=\frac{180-139}{22} \\ \Rightarrow a=1.86\text{ km/s}^2 \end{gathered}[/tex]Conclusion
The acceleration is
[tex]1.86\text{ km./s}^2[/tex]Two spherical objects have masses of 1.5 x 105 kg and 8.5 x 102 kg. Their centers are separated by a distance of 2500 m. Find the gravitational attraction between them.
Given data
*The mass of the first spherical object is m = 1.5 × 10^5 kg
*The mass of the second spherical object is M = 8.5 × 10^2 kg
*The separated distance between the two spherical objects is r = 2500 m
The expression for the gravitational attraction between them is given as
[tex]F=\frac{GmM}{r^2}[/tex]*Here G = 6.67 × 10^-11 N.m^2/kg^2 is the value of the gravitational constant
Substitute the known values in the above expression as
[tex]\begin{gathered} F=\frac{(6.67\times10^{-11})(1.5\times10^5)(8.5\times10^2)}{(2500)^2} \\ =1.36\times10^{-9}\text{ N} \end{gathered}[/tex]Hence, the gravitational attraction between them is F = 1.36 × 10^-9 N