Including Earth and Space Sciences Homework - Assignment Example

Number: Physics: Including Earth and Space sciences Homework Exit Questions 1. Unit of voltage is V 2. Unit of force is Newton N 1.3. Unit of charge is coulomb C 1.4. Unit of current is amperes A 1.5. Unit of resistance is Ohms Ω 1.6. Unit of work, energy in the MKS system of units id Joule J 1.7. 136 volts = 1.36e-16 ev 1.8. 20000ev= 3.20435466e-15 J [1ev = 1.60217657 x 10-19joules or 1.60217733 e-19 J] 1.9. The energy in an electron falling through a potential difference of 15 Volts 1.10. Current I = = 0.5 A 2. Practice Homework 2.1. Resistors of 10, 20 and 30 Ohms connected in a series of a battery 12 volts 2.1.1. Equivalent resistance Rt=R1+R2+R3 Thus, Rt= 10+20+30= 60 Ω 2.1.2. Current in the circuit I= Thus = 0.2 A 2.2. Resistors of 10, 20 and 30 Ohms connected in parallel of a battery 12 volts 2.2.1. Equivalent Resistance = Thus = Thus Rt== 5.45 Ω 2.2.2. Current in the Circuit I= = 2.2 A 3. Priming Homework 3.1. Identify the symbols in qV=mv2, m is mass of electron, v is velocity of electron after exiting a charge and q ionic charge of electron 3.2. Solve for K F=(Kq1*) F*= Kq1 Thus K= [F*)] * (q1)-1 3.3. When the word electron and proton is mentioned there must be provided with an applied voltage 3.4. Measurements in this course should be calculated in Meters m thus convert cm to m. 3.5. The equation used with voltage and velocity is the qV=mv2 3.6. Units used to calculate electricity bill is the kilowatt-hour kWh 3.7. Approximate cost of one u it of electricity is 15 cents/kWh 3.8. Unit of energy is Joules J 3.9. = Power 3.10. Typical Rating of a light bulb in watts Vary from 20 watts, 40 watts, 60 watts, 75 watts, 100 watts. 3.11. Typical Watts of a hair drier 1200-1875 Watts Current drawn by a typical hair drier that uses a voltage of 120V P=V*I thus = 10A = 15.625 A 3.12. Thermostat setting in winter 200C (680F) 3.13. Thermostat setting in summer 780 F to 850F when away from home 3.14. The long a shower I usually take is 5 minutes or 0.12hrs 3.15. Capacitor can be termed as components that store electric charge and have several uses such as Timer circuits, Suppression and coupling, Power factor correction, power conditioning, Motor starters, signal processing. 1. Review of Basic Math skills and concepts 1.1. 0.0001m2= 1 cm2 1.2. 0.000001m3= 1 cm3 1.3. 1,000,000 cm3=1 m3 1.4. 10, 000cm2= 1 m2 1.5. 1,000,000µC = 1 C 1.6. 1C= 1,000,000µC 1.7. 1m= 1000mm 1.8. 1000mm=100cm 2. Review of Course Skills and concepts 2.1. Electron volt is a unit of energy approximately equal to 1.60217657 x 10-19 joules that is the amount of energy gained or lost by the charge of a single (solitary) electron enthused and moved across an electric potential difference of one volt. 2.2. Trigger words for the use of centripetal force equation will include movement in a circular path as well as the presence of a mass of electron 2.3. Units of distance is meters m, speed and acceleration is meters per second and meter per square second respectively, in the MKS system of units 2.4. Unit of force and energy in the MKS system of units 2.5. Unit of charge q is coulomb C 2.6. Ohm’s Law: The current through a conductor between two points is directly proportional to the, pd), potential difference across the two points 2.7. Unit of area is square meters m2 2.8. Unit of Volume is cubic meters m3 2.9. Relationship between I and R in Ohms Law I and R have and inverse relationship that is I= 1/R and R =1/V 2.10. Equation for resistors in circuit Series Rt=R1+R2+R3 and Parallel 3. 4. Class Problems 4.1. Equivalent resistance in the following; battery is 11V as well as I and I1 -1 4.5+ [-1= 5.5Ω I= I= = 2A I1= since V2=Vt-V1 thus V1=2A x 4.5Ω thus 9V therefore V2= 11-9V = 2 V I1= = 1A 4.2. Equivalent resistance and the currents I and I1 Each resistance is 4Ω and voltage is 11Volts Resistance will be thus 3Ω The total resistance will be 4+3+4= 11Ω Current I will be = 1A I1= where V2 = Vt-V1-V3 V1= 1*4 = 4V V3=1*4= 4V Thus V2= 3 volt Thus I1= = 0.75A 1. Exit questions 1.1. 1V= 1e-18 ev 1.2. 236V= 2.36e-16ev 1.3. 1ev=1.60217733 e-19 J 1.4. 100ev= 1.60217733 e -17J 1.5. 1J= 6.241506363e18ev 1.6. 64J=3.999445640272e20 ev 1.7. Equivalent resistance in series circuits Rt= 10+15+20 = 45Ω 1.8. Equivalent resistance in a parallel circuit Rt=[-1 Rt= 54.55Ω 2. Practice Homework 2.1. Equivalent Resistance of The Circuit battery is 12V as well as calculate I and I1 -1 5+ [-1= 6Ω I= I= = 2A I1= since V2=Vt-V1 thus V1=2A x 5Ω thus 10V therefore V2= 12-9V = 2 V I1= = 1.5A 2.2. Equivalent Resistance and Currents I and I1 Each resistance is 6Ω and voltage is 16.5V Simplifying Resistance will be thus 4.5Ω The total resistance will be 6+4.5+6=16.5Ω Current I will be = 1A I1= where V2 = Vt-V1-V3 V1= 1*6 = 6V V3=1*6= 6VThus V2= 4.5volt Thus I1= = 0.75A 3. Priming Homework 3.1. Unit of energy I Joules J 3.2. = power 3.3. Unit of power is watts W 3.4. Typical rating in watts of a bulb is 14 watts – 100 watts 3.5. 100 watts bulb for 10 hours a day in 30 days = 30 kWh * $0.15 = $4.5 3.6. Typical current hair drier draws when on 120 volts US P=V*I thus = 10A = 15.625 A 3.7. Typical current hair drier draws when on 240 volts overseas = 5A = 7.8125A 3.8. I often dry my clothes once a week in a drier 3.9. I use the dish washer in the evening [1800hrs] to clean my dishes once a day 3.10. My shower takes up to 5 minutes or 0.12 hours in a cold tap shower 3.11. It could be possible to reduce but no way not unless I am running late to something 3.12. No. No one has ever told me to turn off the bulb actually I use the energy saving bulb rating 14watts 3.13. Typical monthly electricity bill in my house is 70kWh* 15 cents/kWh 1050 cents dollar 3.14. Capacitor can be termed as components that store electric charge and have several uses such as Timer circuits, Suppression and coupling, Power factor correction, power conditioning, Motor starters, signal processing. 3.15. Ac power can only be used through a help of an AC to DC convertor to charge car batteries or Phone batteries. Answer all questions section: 1. U nit of distance is meters m 2. Unit of speed meters per second ms-2 3. Unit of force is Newton N 4. Unit of energy is Joule J 5. Coulombs Laws in Mathematical form And Meaning of each symbol is a vector equation scalar form |F|= ke Vector form F1=ke ѓ21 where ke is coulombs constant q1 and q2 are the signed magnitudes for the charges .The scalar r is the distance between the charges. The vector r21 = r1-r2 is vectoral distance between the charges and ѓ21is r21/|r21|(a unit vector pointing from q2 to q1) F is the centripetal force 6. Units of constant K in the problem is N.m2.C-2 7. In coulombs law there will be net effects on the force if the magnitude of the two charges and the value of the distance between them are doubled equivalent to 8. In coulombs law there will be net effects on the force if the magnitude of the two charges and the value of the distance between them are doubled equivalent to 9. Electric field in mathematical form and meaning of each symbol 10. If Force F and acceleration a are directly related The proportional relationship is that F=a Equation for the relationship is= 11. Y is inversely related to z and is directly proportional to p, The relationship will be y=1/z and y=p Equation will be y=p/z 12. Acceleration falls through a potential difference of 87 volts. How much energy in electron volts will it gain as a result of this fall? 13. Proton falls through a PD of 916 volts. . How much energy in electron volts will it gain as a result of this fall? 14. In Ohm’s law current I=V/R which of the statements is true a. The current is directly proportional to voltage and inversely proportional to resistance- true. b. The current is inversely proportional to voltage and directly proportional to resistance- false 15. Equivalent resistance in: a. Series circuit Rt= 10+15+30 = 55Ω b. Parallel Circuit Rt=[]-1= 5Ω 16. Identify the symbols Fe= (mv2/R) 17. Circle in a problem requires the area equation 18. The outlets in a power strip are wired in parallel 19. Electric companies have to transmit power at high voltage to reduce energy loss in long distance transmission thus voltage is raised to 110kV or above. 20. 100 watts bulb used for about 10 hours per night would have 30kWh by 8 cents would add up to 240cents $ 2.4 dollars thus B, $10 21. The word electron in a problem physical quantities will be Amperes A 22. A change of +q is placed at a distance of 3 cm from a negative charge, -Q, A second charge of –q is then placed next the first charge, The force –Q charge will: Increase to twice its former magnitude as the positive charge will repel each other 23. Two charges one positive and the other negative are initially 3cm apart. They are then pulled away from each other until they are 9cm away from each other. The force between them is now smaller by a factor of √3 24. Potential difference is a Vector quantity in that we measures it across a medium Note: Answer all Questions Instructions For definitions provide mathematical definition as well as meaning of symbols used in the equation. 1. a. Coulombs law and give the unit of constant in the MKS system of units Scalar form |F|= ke Vector form F1=ke ѓ21 where ke is coulombs constant, q1 and q2 are the signed magnitudes of the charges. The scalar r is the distance between the charges the vector r21 = r1-r2 is vectoral distance between the charges and ѓ21is r21/|r21| (a unit vector pointing from q2 to q1) F is the centripetal force b. Concept of electron volt Electron volt is a unit of energy approximately equal to 1.60217657 x 10-19joules. That is the amount of energy gained, or lost by the charge of a single electron moved through an electric potential difference of one volt. c. Three positive points lie in a straight line, each separated by distance of 40cm. The center charge is 5µC, while one end charge is 3µC, and the other end charge is 8µC. Find the Resultant force on the 8µC charge Given 1µC=10-6C 2. Three 6µC point charges lie on a straight line separated by distances of 60cm. Find the resultant force on the center negative charge if one end charge is negative and the other end charge is positive. 3. a. Ohms Law and provide the unit of resistance I=V/R where: I is the current through a conductor Amperes, V voltage or potential difference measured across a conductor and R is the resistance of the conductor in Ohms b. Current and provide its unit I=V/R where its units is amperes c. Equivalent resistance as well as calculate I and I1 from the battery of 12 volts -1 5.5+ [-1= 6.6Ω I= I= = 1.846A I1= since V2=Vt-V1 thus V1=1.846A x 5.5Ω thus 10.153V therefore V2= 12-9V = 1.847 V I1= = 0.924A 4. a. Power and provide units Energy/Time its units is Watts W b. The resistance of a tungsten wire used in a filament of 100W bulb is 280 ohms when the bulb is hot. Assuming the temperature of the hot filament is to be 21200Cwhat would be the resistance at 200C? 21200C=280 ohms Thus 20*280/2120 200C= 2.64Ω c. We want to operate a 1, 400W dryer, a 400W washer, Three 75 W bulbs and a 60W radio, all from the same 120V line, what would be the minimum rating for such a line power = 400+400+75+60 = 935W and V2/p= R thus Resistance 15.4 I= 120/15.4 7.79A thus the minimum rating should be 10A Bonus Question Calculate the resistance and the current I and I1 in the figure below. Each resistance of 4Ω and the voltage is 11 volts. Resistance will be thus 3Ω The total resistance will be 4+3+4= 11Ω Current I will be = 1A I1= where V2 = Vt-V1-V3 V1= 1*4 = 4V V3=1*4= 4V Thus V2= 3 volt Thus I1= = 0.75A Read More