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Example (Gravitational Force)

The mass of the sun is 2 × 10 30 kg size 12{2 times "10" rSup { size 8{"30"} } ` ital "kg"} {} . The mass of the earth is 5 . 97 × 10 24 kg size 12{5 "." "97" times "10" rSup { size 8{"24"} } ital "kg"} {} . The approximate distance between the sun and the earth is 149 . 6 × 10 6 km . size 12{"149" "." 6 times "10" rSup { size 8{6} } ital "km" "." } {}

Question: What is the magnitude of the gravitational force that the sun exerts on the earth? (Express the result in scientific notation.)

Solution:

F = ( 6 . 67 × 10 − 11 N − m 2 / kg 2 ) ( 2 × 10 30 kg ) ( 5 . 97 × 10 24 kg ) / ( 149 . 6 × 10 6 km ) 2 size 12{F= \( 6 "." "67"` times `"10" rSup { size 8{ - "11"} } `N - m rSup { size 8{2} } / ital "kg" rSup { size 8{2} } \) ` \( 2 times "10" rSup { size 8{"30"} } ital "kg" \) \( 5 "." "97" times "10" rSup { size 8{"24"} } ital "kg" \) /` \( "149" "." 6 times "10" rSup { size 8{6} } ital "km" \) rSup { size 8{2} } } {}
F = ( 6 . 67 × 10 − 11 N − m 2 / kg 2 ) ( 2 × 10 30 kg ) ( 5 . 97 × 10 24 kg ) / ( 149 . 6 × 10 9 m ) 2 size 12{F= \( 6 "." "67"` times `"10" rSup { size 8{ - "11"} } `N - m rSup { size 8{2} } / ital "kg" rSup { size 8{2} } \) ` \( 2 times "10" rSup { size 8{"30"} } ital "kg" \) \( 5 "." "97" times "10" rSup { size 8{"24"} } ital "kg" \) /` \( "149" "." 6 times "10" rSup { size 8{9} } m \) rSup { size 8{2} } } {}
F = ( 6 . 67 × 10 − 11 N − m 2 ) ( 2 × 10 30 ) ( 5 . 97 × 10 24 ) / ( 149 . 6 × 10 9 m ) 2 size 12{F= \( 6 "." "67"` times `"10" rSup { size 8{ - "11"} } `N - m rSup { size 8{2} } \) ` \( 2 times "10" rSup { size 8{"30"} } \) \( 5 "." "97" times "10" rSup { size 8{"24"} } \) /` \( "149" "." 6 times "10" rSup { size 8{9} } m \) rSup { size 8{2} } } {}
F = ( 6 . 67 × 10 − 11 N − m 2 ) ( 2 × 10 30 ) ( 5 . 97 × 10 24 ) / ( 22 , 380 × 10 18 m 2 ) size 12{F= \( 6 "." "67"` times `"10" rSup { size 8{ - "11"} } `N - m rSup { size 8{2} } \) ` \( 2 times "10" rSup { size 8{"30"} } \) \( 5 "." "97" times "10" rSup { size 8{"24"} } \) /` \( "22","380" times "10" rSup { size 8{"18"} } m rSup { size 8{2} } \) } {}
F = ( 6 . 67 × 10 − 11 N ) ( 2 × 10 30 ) ( 5 . 97 × 10 24 ) / ( 22 , 380 × 10 18 ) size 12{F= \( 6 "." "67"` times `"10" rSup { size 8{ - "11"} } `N \) ` \( 2 times "10" rSup { size 8{"30"} } \) \( 5 "." "97" times "10" rSup { size 8{"24"} } \) /` \( "22","380" times "10" rSup { size 8{"18"} } \) } {}
F = ( 6 . 67 × 2 × 5 . 97 ) × ( 10 − 11 × 10 30 × 10 24 ) N / ( 2 . 238 × 10 22 ) size 12{F= \( 6 "." "67"` times `2 times 5 "." "97" \) times \( "10" rSup { size 8{ - "11"} } times "10" rSup { size 8{"30"} } times "10" rSup { size 8{"24"} } \) `N/` \( 2 "." "238" times "10" rSup { size 8{"22"} } \) } {}
F = ( 79 . 64 × 10 43 ) N / ( 2 . 238 × 10 22 ) size 12{F= \( "79" "." "64" times "10" rSup { size 8{"43"} } \) `N/` \( 2 "." "238" times "10" rSup { size 8{"22"} } \) } {}
F = ( 79 . 64 / 2 . 238 ) × 10 43 − 22 N size 12{F= \( "79" "." "64"/2 "." "238" \) times "10" rSup { size 8{"43" - "22"} } `N} {}
F = ( 35 . 6 ) × 10 21 N size 12{F= \( "35" "." 6 \) times "10" rSup { size 8{"21"} } `N} {}
F = 3 . 56 × 10 22 N size 12{F=3 "." "56" times "10" rSup { size 8{"22"} } `N} {}

Coulomb’s law

Electrons and protons are examples of charged particles. In the case of the proton, charge is positive. The charge of a proton is thus said to have positive polarity. On the other hand, an electron has a negative charge. Thus the charge of an electron is said to have negative polarity.

An electrostatic force exists between two charged particles. If the charges associated with the two particles are of the same polarity, the electrostatic force will be repulsive . On the other hand if the charges for two particles have opposite polarity, the electrostatic force will be attractive .

Through extensive laboratory work, the physicist Charles Coulomb first established a mathematical expression that calculates the magnitude of the electrostatic force that results from the interaction of two charged particles. This expression which is known as Coulomb’s law for electrostatic forces is given by

F = k q Q r 2 size 12{F=k` left ( { {q`Q} over {r rSup { size 8{2} } } } right )} {}

where q and Q are the values of the charges measured in the units Coulombs, r is the distance measured in meters that separates the charged particles and k is a constant

k = 9 × 10 9 Newton − meters 2 / Coulomb 2 size 12{k=9` times `"10" rSup { size 8{9} } ` ital "Newton" - ital "meters" rSup { size 8{2} } / ital "Coulomb" rSup { size 8{2} } } {}

It is interesting to note the similarity of the general form of Coulomb’s law for electrostatic forces with Newton’s law of gravitational force. Just as was the case with the gravitational force, the magnitude of the electrostatic force decreases with the square of the distance separating the two particles.

Example (Electrostatic Force)

The charge on an electron is given by the equation

q e = − 1 . 60 × 10 − 19 Coulombs ( C ) size 12{q rSub { size 8{e} } = - 1 "." "60" times "10" rSup { size 8{ - "19"} } ital "Coulombs"` \( C \) } {}

The charge associated with a proton is

q p =+ 1 . 60 × 10 − 19 Coulombs ( C ) size 12{q rSub { size 8{p} } "=+"1 "." "60" times "10" rSup { size 8{ - "19"} } ital "Coulombs"` \( C \) } {}

Suppose that an electron and a proton are separated by a distance of 10 − 6 nanometers . size 12{"10" rSup { size 8{ - 6} } ital "nanometers" "." } {}

Questions: What is the magnitude of the electrostatic force that the protonexerts on the electron? Is the force attractive or repulsive?

Solution:

F = k q e q p r 2 size 12{F=k` { {q rSub { size 8{e} } `q rSub { size 8{p} } } over {r rSup { size 8{2} } } } } {}
F = ( 9 × 10 9 N − m 2 / C 2 ) ( − 1 . 60 × 10 − 19 C ) ( 1 . 60 × 10 − 19 C ) ( 10 − 6 × nanometers ) 2 size 12{F= { { \( 9` times `"10" rSup { size 8{9} } `N - m rSup { size 8{2} } /C rSup { size 8{2} } \) ` \( - 1 "." "60" times "10" rSup { size 8{ - "19"} } C \) \( 1 "." "60" times "10" rSup { size 8{ - "19"} } C \) } over { \( "10" rSup { size 8{ - 6} } times ital "nanometers" \) rSup { size 8{2} } } } } {}
F = ( 9 × 10 9 ) ( − 1 . 60 × 10 − 19 ) ( 1 . 60 × 10 − 19 ) N − m 2 ( 10 − 6 × 10 − 9 m ) 2 size 12{F= { { \( 9` times `"10" rSup { size 8{9} } \) ` \( - 1 "." "60" times "10" rSup { size 8{ - "19"} } \) \( 1 "." "60" times "10" rSup { size 8{ - "19"} } \) `N - m rSup { size 8{2} } } over { \( "10" rSup { size 8{ - 6} } times "10" rSup { size 8{ - 9} } `m \) rSup { size 8{2} } } } } {}
F = ( 9 ) ( − 1 . 60 ) ( 1 . 60 ) ( 10 9 × 10 − 19 × 10 − 19 ) N − m ( 10 − 15 m ) 2 size 12{F= { { \( 9` \) ` \( - 1 "." "60" \) \( 1 "." "60" \) ` \( "10" rSup { size 8{9} } times "10" rSup { size 8{ - "19"} } times "10" rSup { size 8{ - "19"} } \) N - m} over { \( "10" rSup { size 8{ - "15"} } `m \) rSup { size 8{2} } } } } {}
F = ( 23 . 0 × 10 − 29 ) N ( 10 − 15 ) 2 size 12{F= { { \( "23" "." 0` times "10" rSup { size 8{ - "29"} } \) N} over { \( "10" rSup { size 8{ - "15"} } \) rSup { size 8{2} } } } } {}
F = 23 . 0 × 10 − 29 × 10 30 N size 12{F="23" "." 0 times "10" rSup { size 8{ - "29"} } times "10" rSup { size 8{"30"} } `N} {}
F = 23 . 0 × 10 1 N size 12{F="23" "." 0 times "10" rSup { size 8{1} } `N} {}
F = 230 N size 12{F="230"`N} {}

Because the charges on the electron and the proton differ in polarity, the electrostatic force is attractive .

Exercises

  1. The weight of the Space Shuttle is 4,470,000 lb. Express this weight in scientific notation.
  2. The weight of a honey bee is 0.000 385 05 lb. Express this weight in scientific notation.
  3. Perform the following multiplications. (a) (37.5 x 10 7 ) x (2.87 x 10 5 ), (b) (37.5 x 10 7 ) x (2.87 x 10 -5 ).
  4. Perform the following divisions. (a) (37.5 x 10 7 ) / (2.87 x 10 5 ), (b) (37.5 x 10 7 ) / (2.87 x 10 -5 ).
  5. The area of the United States is 9.83 x 10 6 km 2 . Its population is 3.10 x 10 6 . What is the population density (people/km 2 ) of the United States?
  6. The distance between the Sun and the Earth is 1.47 x 10 11 m. The speed of light is 299,792,458 m/s. How long does it take for light to travel between the Sun and the Earth?
  7. The planned Blythe Solar Power Plant in California is expected to produce 968 MW. The Aswan Dam Power Plant in Egypt produces 2.1 gigawatts of power. How may solar power plants with similar power production as that of the Blythe Solar Power Plant would be needed to match the power production of the Aswan Dam Power Plant?
  8. An electric circuit consists of a 9.00 V battery in series with a load that has a resistance of 13.78 kΩ. (a) Find the power delivered by the battery. (b) Find the power absorbed by the load.
  9. The mass of the Earth is 5.97 x 10 24 kg, while that of the moon is 7.36 x 10 22 kg. The average distance between the Earth and the Moon is 384,403 km. What is he force exerted on the Moon by the Earth?
  10. Two charges are separated by 4.66 x 10 -8 m. Each has a positive charge of 3.6 x 10 -18 Coulombs. What is the repulsive force that they exert on one another?

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
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David Reply
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emma Reply
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what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
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Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Magreth
progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
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Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
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