Are you over 18 and want to see adult content?
More Annotations
A complete backup of sutherlandinstitute.org
Are you over 18 and want to see adult content?
A complete backup of newsamericasnow.com
Are you over 18 and want to see adult content?
A complete backup of rikertnordic.com
Are you over 18 and want to see adult content?
A complete backup of simmonsfirst.com
Are you over 18 and want to see adult content?
A complete backup of librosaguilar.com
Are you over 18 and want to see adult content?
A complete backup of altenpflege-messe.de
Are you over 18 and want to see adult content?
Favourite Annotations
A complete backup of canyoneeringnorthwest.com
Are you over 18 and want to see adult content?
A complete backup of wallpaperama.com
Are you over 18 and want to see adult content?
A complete backup of honestburgers.co.uk
Are you over 18 and want to see adult content?
A complete backup of muvendis.wordpress.com
Are you over 18 and want to see adult content?
Text
Original length (l
COULOMB’S LAW
1. Two point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude of the electric force. The constant k = 8.988 x 109 Nm2C Advertisement −2 = 9 x 109 Nm2C−2. Known : Charge A (qA) = +8 μC = +8LINEAR EXPANSION
1. A steel is 40 cm long at 20 oC. The coefficient of linear expansion for steel is 12 x 10-6 (Co)-1. The increase in length and the final length when it is at 70 oC will be Known : The change in temperature (ΔT) = 70oC – 20oC = 50oC The original length (L1) WORK-MECHANICAL ENERGY PRINCIPLE 1. A 2-kg box decelerated from 10 m/s to rest. The coefficient of kinetic friction is 0.2. Acceleration due to gravity is 10 m/s2. What is the magnitude of displacement? Known ; Mass (m) = 2 kg Initial velocity (vo) = 10 m/s Final velocity (vt) = 0 m/s The coefficient of kinetic friction (μk)BUOYANT FORCE
Related Posts. Force of gravity and gravitational field – problems and solutions. 1. Two objects m1 and m2 each with a mass of 6 kg and 9 kg separated by a distance of 5 NEWTON’S SECOND LAW OF MOTION Solved problems in Newton’s laws of motion – Newton’s second law of motion 1. A 1 kg object accelerated at a constant 5 m/s2. Estimate the net force needed to accelerate the object. Known : Mass (m) = 1 kg Acceleration (a) = 5 m/s2 Wanted : net force (∑F) Solution : We useNewton’s
PEMBAHASAN SOAL GELOMBANG BUNYI 1. Soal UN 2008/2009 P04 No.24 Seorang penonton pada lomba balap mobil mendengar bunyi (deru mobil) yang berbeda, ketika mobil mendekat dan menjauh. Rata-rata mobil balap mengeluarkan bunyi 800 Hz. Jika kecepatan bunyi di udara 340 m.s-1 dan kecepatan mobil 20 m.s-1, maka frekuensi yang di dengar saat mobil mendekat adalah. A. 805 Hz B.ELECTRIC FIELD
1. Point A located at the center between two charges. Both charges have the same magnitude but opposite sign and separated by a distance of a. The magnitude of the electric field at point A is 36 N/C. If point A moved 1/2a close to one of both charges, what is the magnitudeof the
CARNOT CYCLE
1. If heat absorbed by the engine (Q1) = 10,000 Joule, what is the work done by the Carnot engine? Known: Low temperature (T2) = 400 K High temperature (T1) = 800 K Heat input (Q1 Advertisement Advertisement ) = 10,000 Joule Wanted: Work done by Carnot engine EQUATION OF DIVERGING (CONCAVE) LENS Based on the the the sign rules of the concave lens, this equation can be changed to like the equation of curved mirror, if the image distance (di) is given a negative sign because the beam of light does not pass the image and the focal length (f) is also given a negative sign because the focal point of the concave lens is not passed by light (compare with the figure of the image formation above). STRESS, STRAIN, YOUNG’S MODULUS 2. A cord has original length of 100 cm is pulled by a force. The change in length of the cord is 2 mm. Determine the strain! Known :Original length (l
COULOMB’S LAW
1. Two point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude of the electric force. The constant k = 8.988 x 109 Nm2C Advertisement −2 = 9 x 109 Nm2C−2. Known : Charge A (qA) = +8 μC = +8LINEAR EXPANSION
1. A steel is 40 cm long at 20 oC. The coefficient of linear expansion for steel is 12 x 10-6 (Co)-1. The increase in length and the final length when it is at 70 oC will be Known : The change in temperature (ΔT) = 70oC – 20oC = 50oC The original length (L1) WORK-MECHANICAL ENERGY PRINCIPLE 1. A 2-kg box decelerated from 10 m/s to rest. The coefficient of kinetic friction is 0.2. Acceleration due to gravity is 10 m/s2. What is the magnitude of displacement? Known ; Mass (m) = 2 kg Initial velocity (vo) = 10 m/s Final velocity (vt) = 0 m/s The coefficient of kinetic friction (μk)BUOYANT FORCE
Related Posts. Force of gravity and gravitational field – problems and solutions. 1. Two objects m1 and m2 each with a mass of 6 kg and 9 kg separated by a distance of 5 NEWTON’S SECOND LAW OF MOTION Solved problems in Newton’s laws of motion – Newton’s second law of motion 1. A 1 kg object accelerated at a constant 5 m/s2. Estimate the net force needed to accelerate the object. Known : Mass (m) = 1 kg Acceleration (a) = 5 m/s2 Wanted : net force (∑F) Solution : We useNewton’s
PEMBAHASAN SOAL GELOMBANG BUNYI 1. Soal UN 2008/2009 P04 No.24 Seorang penonton pada lomba balap mobil mendengar bunyi (deru mobil) yang berbeda, ketika mobil mendekat dan menjauh. Rata-rata mobil balap mengeluarkan bunyi 800 Hz. Jika kecepatan bunyi di udara 340 m.s-1 dan kecepatan mobil 20 m.s-1, maka frekuensi yang di dengar saat mobil mendekat adalah. A. 805 Hz B.ELECTRIC FIELD
1. Point A located at the center between two charges. Both charges have the same magnitude but opposite sign and separated by a distance of a. The magnitude of the electric field at point A is 36 N/C. If point A moved 1/2a close to one of both charges, what is the magnitudeof the
CARNOT CYCLE
1. If heat absorbed by the engine (Q1) = 10,000 Joule, what is the work done by the Carnot engine? Known: Low temperature (T2) = 400 K High temperature (T1) = 800 K Heat input (Q1 Advertisement Advertisement ) = 10,000 Joule Wanted: Work done by Carnot engine EQUATION OF DIVERGING (CONCAVE) LENS Based on the the the sign rules of the concave lens, this equation can be changed to like the equation of curved mirror, if the image distance (di) is given a negative sign because the beam of light does not pass the image and the focal length (f) is also given a negative sign because the focal point of the concave lens is not passed by light (compare with the figure of the image formation above). NEWTON’S SECOND LAW OF MOTION Solved problems in Newton’s laws of motion – Newton’s second law of motion 1. A 1 kg object accelerated at a constant 5 m/s2. Estimate the net force needed to accelerate the object. Known : Mass (m) = 1 kg Acceleration (a) = 5 m/s2 Wanted : net force (∑F) Solution : We useNewton’s
POWER – PROBLEMS AND SOLUTIONS 1. A 50-kg person runs up the stairs 10 meters high in 2 minutes. Acceleration due to gravity (g) is 10 m/s2. Determine the power. Known : Mass (m) = 50 kg Height (h) = 10 meters Acceleration due to gravity (g) = 10 m/s2 Time interval (t) = 2 minute = 2 (60) =FREE FALL MOTION
Free fall motion – problems and solutions. 1. A stone free fall from the height of 45 meters. If the acceleration due to gravity is 10 ms-2, what is the speed of the stone when hits the ground?. Known : Height (h) = 45 meters. Acceleration due to gravity (g) = 10 m/s 2. Wanted : The final velocity of the stone when it hits the ground (v t)Solution :
AVERAGE SPEED AND AVERAGE VELOCITY Solved Problems in Linear Motion – Average speed and average velocity 1. A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second. Determine average speed and average velocity. Solution Distance = 100 meters + 50 meters = 150 meters DisplacementELECTRIC FIELD
1. Point A located at the center between two charges. Both charges have the same magnitude but opposite sign and separated by a distance of a. The magnitude of the electric field at point A is 36 N/C. If point A moved 1/2a close to one of both charges, what is the magnitudeof the
CONSERVATION OF MECHANICAL ENERGY 2. An ice skier sliding from a height of A, as shown in the figure below. If the initial velocity = 0 and acceleration due to gravity is 10 ms-2, then what is the velocity of the skier at point B.. Known : Initial velocity (v o) = 0. Acceleration due to gravity (g) = 10 m/s 2. The change in height = 50 meters – 10 meters = 40 metersINTENSITY OF SOUND
Intensity of sound – problems and solutions. 1. Point A and B located at 4 meters and 9 meters from a source of the sound. If I A and I B are intensity at point A and point B, then I A: I B =. Known : The distance of point A from a source of sound (r A) = 4 meters. A distance of point B from the source of sound (r B) = 9 meters. The intensity of sound at point A = I APROJECTILE MOTION
Projectile motion – problems and solutions. 1. A bullet fired a t an angle θ = 60 o with a velocity of 20 m/s. Acceleration due to gravity is 10 m/s 2.What is the time interval to reach the maximum height? DISTANCE AND DISPLACEMENT Solved Problems in Linear Motion – Distance and displacement 1. A car travels along a straight road 100 m east then 50 m west. Find distance and displacement of the car. Solution Distance is 100 meters + 50 meters = 150 meters Displacement is 100 meters – 50 meters = 50 meters, to the east.CONVEX MIRROR
1. The focal length of a convex mirror is 10 cm and the object distance is 20 cm. Determine (a) the image distance (b) the magnification of image Known : The focal length (f) = -10 cm The minus sign indicates that the focal point of convex mirror is virtual The object distance (do) = STRESS, STRAIN, YOUNG’S MODULUS A string has a diameter of 1 cm and the original length of 2 m. The string is pulled by a force of 200 N. Determine the change in length of the string! Young’s modulus of the string = 5 x 109 N/m2. Known : Young’s modulus (E) = 5 x 109 N/m2. Original length (l0) = NEWTON’S SECOND LAW OF MOTION Solved problems in Newton’s laws of motion – Newton’s second law of motion 1. A 1 kg object accelerated at a constant 5 m/s2. Estimate the net force needed to accelerate the object. Known : Mass (m) = 1 kg Acceleration (a) = 5 m/s2 Wanted : net force (∑F) Solution : We useNewton’s
WORK DONE BY FORCE
Work done by force – problems and solutions. 1. A person pulls a block 2 m along a horizontal surface by a constant force F = 20 N. Determine the work done by force F acting on the block. Known : Force (F) = 20 N. Displacement (s) = 2 m. Angle (θ) = 0. Wanted : Work (W)Solution :
COULOMB’S LAW
Coulomb’s law – problems and solutions. 1. Two point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude of the electric force. The constant k = 8.988 x 109 Nm2C. Advertisement. Advertisement. −2 = 9 x 109 Nm2C−2. WORK-MECHANICAL ENERGY PRINCIPLE Work-mechanical energy principle – problems and solutions. 1. A 2-kg box decelerated from 10 m/s to rest. The coefficient of kinetic friction is 0.2. Acceleration due to gravity is 10 m/s2.BUOYANT FORCE
Wanted : Density of gold. Solution : Weight of gold in air : w = m g = ρb V g —– Equation 1. V is volume of gold in kerosene. Buoyant force (FB) equal to the weight of gold in air (w) minus weight of gold in kerosene (w’) : w – w’ = FA. w – w’ = ρf V g —– Equation 2. V is the volume of gold in kerosene. MOTION ON THE INCLINED PLANE WITHOUT THE FRICTION FORCE Motion on the inclined plane without the friction force – application of Newton’s law of motion problems and solutions. 1. Box’s mass = 2 kg, acceleration due to gravity = 9.8 m/s 2.Find (a) the net force which accelerates the box downward (b) magnitude of the box’s acceleration.. SolutionINTENSITY OF SOUND
Solution : The addition of the sound intensity of 10 W/m2 = 10-3 W/cm2 is equivalent to the addition of the sound intensity level of 10 dB. The addition of sound intensity of 102 W/m2 = 10-2 W/cm2 is equivalent to the addition of sound intensity level of 20 dB. And so on. If the intensity level is increased by 10 dB, the intensity increases by AVERAGE SPEED AND AVERAGE VELOCITY Solved Problems in Linear Motion – Average speed and average velocity 1. A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second. Determine average speed and average velocity. Solution Distance = 100 meters + 50 meters = 150 meters DisplacementCARNOT CYCLE
6. An Carnot engine absorbs heat at high temperature 800 Kelvin and efficiency of the Carnot engine is 50%. What is the high temperature to increase efficiency to 80% if the low temperature kept constant. Known : If high temperature (TH) = 800 K , efficiency (e) = 50% = 0.5. Wanted : High temperature (TH) if efficiency (e) = 80% = 0.8. Solution:
STRESS, STRAIN, YOUNG’S MODULUS A string has a diameter of 1 cm and the original length of 2 m. The string is pulled by a force of 200 N. Determine the change in length of the string! Young’s modulus of the string = 5 x 109 N/m2. Known : Young’s modulus (E) = 5 x 109 N/m2. Original length (l0) = NEWTON’S SECOND LAW OF MOTION Solved problems in Newton’s laws of motion – Newton’s second law of motion 1. A 1 kg object accelerated at a constant 5 m/s2. Estimate the net force needed to accelerate the object. Known : Mass (m) = 1 kg Acceleration (a) = 5 m/s2 Wanted : net force (∑F) Solution : We useNewton’s
WORK DONE BY FORCE
Work done by force – problems and solutions. 1. A person pulls a block 2 m along a horizontal surface by a constant force F = 20 N. Determine the work done by force F acting on the block. Known : Force (F) = 20 N. Displacement (s) = 2 m. Angle (θ) = 0. Wanted : Work (W)Solution :
COULOMB’S LAW
Coulomb’s law – problems and solutions. 1. Two point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude of the electric force. The constant k = 8.988 x 109 Nm2C. Advertisement. Advertisement. −2 = 9 x 109 Nm2C−2. WORK-MECHANICAL ENERGY PRINCIPLE Work-mechanical energy principle – problems and solutions. 1. A 2-kg box decelerated from 10 m/s to rest. The coefficient of kinetic friction is 0.2. Acceleration due to gravity is 10 m/s2.BUOYANT FORCE
Wanted : Density of gold. Solution : Weight of gold in air : w = m g = ρb V g —– Equation 1. V is volume of gold in kerosene. Buoyant force (FB) equal to the weight of gold in air (w) minus weight of gold in kerosene (w’) : w – w’ = FA. w – w’ = ρf V g —– Equation 2. V is the volume of gold in kerosene. MOTION ON THE INCLINED PLANE WITHOUT THE FRICTION FORCE Motion on the inclined plane without the friction force – application of Newton’s law of motion problems and solutions. 1. Box’s mass = 2 kg, acceleration due to gravity = 9.8 m/s 2.Find (a) the net force which accelerates the box downward (b) magnitude of the box’s acceleration.. SolutionINTENSITY OF SOUND
Solution : The addition of the sound intensity of 10 W/m2 = 10-3 W/cm2 is equivalent to the addition of the sound intensity level of 10 dB. The addition of sound intensity of 102 W/m2 = 10-2 W/cm2 is equivalent to the addition of sound intensity level of 20 dB. And so on. If the intensity level is increased by 10 dB, the intensity increases by AVERAGE SPEED AND AVERAGE VELOCITY Solved Problems in Linear Motion – Average speed and average velocity 1. A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second. Determine average speed and average velocity. Solution Distance = 100 meters + 50 meters = 150 meters DisplacementCARNOT CYCLE
6. An Carnot engine absorbs heat at high temperature 800 Kelvin and efficiency of the Carnot engine is 50%. What is the high temperature to increase efficiency to 80% if the low temperature kept constant. Known : If high temperature (TH) = 800 K , efficiency (e) = 50% = 0.5. Wanted : High temperature (TH) if efficiency (e) = 80% = 0.8. Solution:
BUOYANT FORCE
Wanted : Density of gold. Solution : Weight of gold in air : w = m g = ρb V g —– Equation 1. V is volume of gold in kerosene. Buoyant force (FB) equal to the weight of gold in air (w) minus weight of gold in kerosene (w’) : w – w’ = FA. w – w’ = ρf V g —– Equation 2. V is the volume of gold in kerosene. POWER – PROBLEMS AND SOLUTIONS Power – problems and solutions. 1. A 50-kg person runs up the stairs 10 meters high in 2 minutes. Acceleration due to gravity (g) is 10 m/s2. Determine the power. W = work, F = force, w = weight, d = displacement, h = height, m = mass, g = acceleration due to gravity. WWORK DONE BY FORCE
Work done by force – problems and solutions. 1. A person pulls a block 2 m along a horizontal surface by a constant force F = 20 N. Determine the work done by force F acting on the block. Known : Force (F) = 20 N. Displacement (s) = 2 m. Angle (θ) = 0. Wanted : Work (W)Solution :
ELECTRIC CIRCUITS
Which one of the electric circuits as shown below has the bigger current. Solution : The resistance of the resistor is R and the electric voltage is V. Answer A. R1, R2 and R3 are connected in series. The equivalent resistor : RA = R1 + R2 + R3 = R + R + R = 3R. Electric current (I) : Answer B.COULOMB’S LAW
Coulomb’s law – problems and solutions. 1. Two point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude of the electric force. The constant k = 8.988 x 109 Nm2C. Advertisement. Advertisement. −2 = 9 x 109 Nm2C−2.LINEAR EXPANSION
1. A steel is 40 cm long at 20 oC. The coefficient of linear expansion for steel is 12 x 10-6 (Co)-1. The increase in length and the final length when it is at 70 oC will be Known : The change in temperature (ΔT) = 70oC – 20oC = 50oC The original length (L1) EQUATION OF DIVERGING (CONCAVE) LENS Based on the the the sign rules of the concave lens, this equation can be changed to like the equation of curved mirror, if the image distance (di) is given a negative sign because the beam of light does not pass the image and the focal length (f) is also given a negative sign because the focal point of the concave lens is not passed by light (compare with the figure of the image formation above). CAPACITORS IN SERIES AND PARALLEL 1. Three capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 4 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination. Known : Capacitor C1 = 2 μF Capacitor C2 = 4 μF Capacitor C3 = 4 μF Wanted TEMPERATURE AND HEAT Temperature and heat – problems and solutions. 1. On a thermometer X, the freezing point of water at -30o and the boiling point of water at 90o. 60OX = .. oC. On the Fahrenheit scale, the freezing point of water is 32oF and the boiling point of water is 212oF. Between the freezing point and the boiling point, 212o – 32o = 180o.PRESSURE OF FLUIDS
Pressure of fluids – problems and solutions. 1. Acceleration due to gravity is 10 N/kg. The surface area of fish pressed by the water above it is 6 cm2. Determine the force of water above fish that acts on fish. 2. The normal pressure of blood is 80 mm hg to 120 mm hg. NEWTON’S SECOND LAW OF MOTION Solved problems in Newton’s laws of motion – Newton’s second law of motion 1. A 1 kg object accelerated at a constant 5 m/s2. Estimate the net force needed to accelerate the object. Known : Mass (m) = 1 kg Acceleration (a) = 5 m/s2 Wanted : net force (∑F) Solution : We useNewton’s
HOOKE’S LAW
1. A graph of force (F) versus elongation (x) shown in the figure below. Find the spring constant! Solution Hooke’s law formula : k = F / x F = force (Newton) k = spring constant (Newton/meter) x = the change in length (meter) Spring constant : k = 10 / 0.02 = 20 /CARNOT CYCLE
6. An Carnot engine absorbs heat at high temperature 800 Kelvin and efficiency of the Carnot engine is 50%. What is the high temperature to increase efficiency to 80% if the low temperature kept constant. Known : If high temperature (TH) = 800 K , efficiency (e) = 50% = 0.5. Wanted : High temperature (TH) if efficiency (e) = 80% = 0.8. Solution:
WORK-MECHANICAL ENERGY PRINCIPLE Work-mechanical energy principle – problems and solutions. 1. A 2-kg box decelerated from 10 m/s to rest. The coefficient of kinetic friction is 0.2. Acceleration due to gravity is 10 m/s2. AVERAGE SPEED AND AVERAGE VELOCITY Solved Problems in Linear Motion – Average speed and average velocity 1. A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second. Determine average speed and average velocity. Solution Distance = 100 meters + 50 meters = 150 meters Displacement KINETIC THEORY OF GASES Kinetic theory of gases – problems and solutions. 1. Ideal gases in a closed container initially have volume V and pressure P. If the final pressure is 4P and the volume is kept constant, what is the ratio of the initial kinetic energy with the final kinetic energy. Known : Initial pressure (P1) = P. DISTANCE AND DISPLACEMENT Solved Problems in Linear Motion – Distance and displacement 1. A car travels along a straight road 100 m east then 50 m west. Find distance and displacement of the car. Solution Distance is 100 meters + 50 meters = 150 meters Displacement is 100 meters – 50 meters = 50 meters, to the east.LINEAR EXPANSION
1. A steel is 40 cm long at 20 oC. The coefficient of linear expansion for steel is 12 x 10-6 (Co)-1. The increase in length and the final length when it is at 70 oC will be Known : The change in temperature (ΔT) = 70oC – 20oC = 50oC The original length (L1) CAPACITORS IN SERIES AND PARALLEL 1. Three capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 4 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination. Known : Capacitor C1 = 2 μF Capacitor C2 = 4 μF Capacitor C3 = 4 μF Wanted PEMBAHASAN SOAL GELOMBANG BUNYI Pembahasan soal gelombang bunyi. 1. Soal UN 2008/2009 P04 No.24. Seorang penonton pada lomba balap mobil mendengar bunyi (deru mobil) yang berbeda, ketika mobil mendekat dan menjauh. Rata-rata mobil balap mengeluarkan bunyi 800 Hz. Jika kecepatan bunyi di udara 340 m.s-1 dan kecepatan mobil 20 m.s-1, maka frekuensi yang di dengar saat mobil NEWTON’S SECOND LAW OF MOTION Solved problems in Newton’s laws of motion – Newton’s second law of motion 1. A 1 kg object accelerated at a constant 5 m/s2. Estimate the net force needed to accelerate the object. Known : Mass (m) = 1 kg Acceleration (a) = 5 m/s2 Wanted : net force (∑F) Solution : We useNewton’s
HOOKE’S LAW
1. A graph of force (F) versus elongation (x) shown in the figure below. Find the spring constant! Solution Hooke’s law formula : k = F / x F = force (Newton) k = spring constant (Newton/meter) x = the change in length (meter) Spring constant : k = 10 / 0.02 = 20 /CARNOT CYCLE
6. An Carnot engine absorbs heat at high temperature 800 Kelvin and efficiency of the Carnot engine is 50%. What is the high temperature to increase efficiency to 80% if the low temperature kept constant. Known : If high temperature (TH) = 800 K , efficiency (e) = 50% = 0.5. Wanted : High temperature (TH) if efficiency (e) = 80% = 0.8. Solution:
WORK-MECHANICAL ENERGY PRINCIPLE Work-mechanical energy principle – problems and solutions. 1. A 2-kg box decelerated from 10 m/s to rest. The coefficient of kinetic friction is 0.2. Acceleration due to gravity is 10 m/s2. AVERAGE SPEED AND AVERAGE VELOCITY Solved Problems in Linear Motion – Average speed and average velocity 1. A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second. Determine average speed and average velocity. Solution Distance = 100 meters + 50 meters = 150 meters Displacement KINETIC THEORY OF GASES Kinetic theory of gases – problems and solutions. 1. Ideal gases in a closed container initially have volume V and pressure P. If the final pressure is 4P and the volume is kept constant, what is the ratio of the initial kinetic energy with the final kinetic energy. Known : Initial pressure (P1) = P. DISTANCE AND DISPLACEMENT Solved Problems in Linear Motion – Distance and displacement 1. A car travels along a straight road 100 m east then 50 m west. Find distance and displacement of the car. Solution Distance is 100 meters + 50 meters = 150 meters Displacement is 100 meters – 50 meters = 50 meters, to the east.LINEAR EXPANSION
1. A steel is 40 cm long at 20 oC. The coefficient of linear expansion for steel is 12 x 10-6 (Co)-1. The increase in length and the final length when it is at 70 oC will be Known : The change in temperature (ΔT) = 70oC – 20oC = 50oC The original length (L1) CAPACITORS IN SERIES AND PARALLEL 1. Three capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 4 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination. Known : Capacitor C1 = 2 μF Capacitor C2 = 4 μF Capacitor C3 = 4 μF Wanted PEMBAHASAN SOAL GELOMBANG BUNYI Pembahasan soal gelombang bunyi. 1. Soal UN 2008/2009 P04 No.24. Seorang penonton pada lomba balap mobil mendengar bunyi (deru mobil) yang berbeda, ketika mobil mendekat dan menjauh. Rata-rata mobil balap mengeluarkan bunyi 800 Hz. Jika kecepatan bunyi di udara 340 m.s-1 dan kecepatan mobil 20 m.s-1, maka frekuensi yang di dengar saat mobilLINEAR MOTION
Linear motion – problems and solutions. 1. Graph of velocity (v) vs. time (t) shown in the figure below. What is the deceleration according to the graph. Solution. A-B = motion at constant acceleration, B-C = motion at constant velocity, C-D = motion at constant deceleration.LINEAR EXPANSION
1. A steel is 40 cm long at 20 oC. The coefficient of linear expansion for steel is 12 x 10-6 (Co)-1. The increase in length and the final length when it is at 70 oC will be Known : The change in temperature (ΔT) = 70oC – 20oC = 50oC The original length (L1) POWER – PROBLEMS AND SOLUTIONS Power – problems and solutions. 1. A 50-kg person runs up the stairs 10 meters high in 2 minutes. Acceleration due to gravity (g) is 10 m/s2. Determine the power. W = work, F = force, w = weight, d = displacement, h = height, m = mass, g = acceleration due to gravity. WOHM’S LAW
Ohm’s law : V = I R or R = V / I. R1 = V / I = 1.50 / 0.08 = 18.75 Ohm. R2 = V / I = 2.80 / 1.50 = 2.87 Ohm. R3 = V / I = 3.99 / 2.10 = 1.9 Ohm. Based on the table, the relationship between V, I and R is the value of R inversely proportional to V and I. That means thelarger the V
INTENSITY OF SOUND
Solution : The addition of the sound intensity of 10 W/m2 = 10-3 W/cm2 is equivalent to the addition of the sound intensity level of 10 dB. The addition of sound intensity of 102 W/m2 = 10-2 W/cm2 is equivalent to the addition of sound intensity level of 20 dB. And so on. If the intensity level is increased by 10 dB, the intensity increases byFLUID DYNAMICS
Fluid dynamics – problems and solutions. Torricelli’s theorem. 1. A container filled with water and there is a hole, as shown in the figure below. If acceleration due to gravity is 10 ms-2, what is the speed of water through that hole?. Known : DENSITY – PROBLEMS AND SOLUTIONS Mass (m) = 300 gram. Volume (V) = volume of spilled water = 20 cm3. Wanted : density. Solution : ρ = m / V = 300 gram / 20 cm3 = 15 gram/cm3. Read : Heat transfer conduction – problems and solutions. 7. Mass of object is 316 gram, placed in a container as shown in figure. What is the density of the object. AVERAGE SPEED AND AVERAGE VELOCITY Solved Problems in Linear Motion – Average speed and average velocity 1. A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second. Determine average speed and average velocity. Solution Distance = 100 meters + 50 meters = 150 meters DisplacementBUOYANT FORCE
Wanted : Density of gold. Solution : Weight of gold in air : w = m g = ρb V g —– Equation 1. V is volume of gold in kerosene. Buoyant force (FB) equal to the weight of gold in air (w) minus weight of gold in kerosene (w’) : w – w’ = FA. w – w’ = ρf V g —– Equation 2. V is the volume of gold in kerosene. CAPACITORS IN SERIES AND PARALLEL 1. Three capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 4 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination. Known : Capacitor C1 = 2 μF Capacitor C2 = 4 μF Capacitor C3 = 4 μF Wanted NEWTON’S SECOND LAW OF MOTION Solved problems in Newton’s laws of motion – Newton’s second law of motion 1. A 1 kg object accelerated at a constant 5 m/s2. Estimate the net force needed to accelerate the object. Known : Mass (m) = 1 kg Acceleration (a) = 5 m/s2 Wanted : net force (∑F) Solution : We useNewton’s
HOOKE’S LAW
1. A graph of force (F) versus elongation (x) shown in the figure below. Find the spring constant! Solution Hooke’s law formula : k = F / x F = force (Newton) k = spring constant (Newton/meter) x = the change in length (meter) Spring constant : k = 10 / 0.02 = 20 /CARNOT CYCLE
6. An Carnot engine absorbs heat at high temperature 800 Kelvin and efficiency of the Carnot engine is 50%. What is the high temperature to increase efficiency to 80% if the low temperature kept constant. Known : If high temperature (TH) = 800 K , efficiency (e) = 50% = 0.5. Wanted : High temperature (TH) if efficiency (e) = 80% = 0.8. Solution:
WORK-MECHANICAL ENERGY PRINCIPLE Work-mechanical energy principle – problems and solutions. 1. A 2-kg box decelerated from 10 m/s to rest. The coefficient of kinetic friction is 0.2. Acceleration due to gravity is 10 m/s2. AVERAGE SPEED AND AVERAGE VELOCITY Solved Problems in Linear Motion – Average speed and average velocity 1. A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second. Determine average speed and average velocity. Solution Distance = 100 meters + 50 meters = 150 meters Displacement KINETIC THEORY OF GASES Kinetic theory of gases – problems and solutions. 1. Ideal gases in a closed container initially have volume V and pressure P. If the final pressure is 4P and the volume is kept constant, what is the ratio of the initial kinetic energy with the final kinetic energy. Known : Initial pressure (P1) = P. DISTANCE AND DISPLACEMENT Solved Problems in Linear Motion – Distance and displacement 1. A car travels along a straight road 100 m east then 50 m west. Find distance and displacement of the car. Solution Distance is 100 meters + 50 meters = 150 meters Displacement is 100 meters – 50 meters = 50 meters, to the east.LINEAR EXPANSION
1. A steel is 40 cm long at 20 oC. The coefficient of linear expansion for steel is 12 x 10-6 (Co)-1. The increase in length and the final length when it is at 70 oC will be Known : The change in temperature (ΔT) = 70oC – 20oC = 50oC The original length (L1) CAPACITORS IN SERIES AND PARALLEL 1. Three capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 4 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination. Known : Capacitor C1 = 2 μF Capacitor C2 = 4 μF Capacitor C3 = 4 μF Wanted PEMBAHASAN SOAL GELOMBANG BUNYI Pembahasan soal gelombang bunyi. 1. Soal UN 2008/2009 P04 No.24. Seorang penonton pada lomba balap mobil mendengar bunyi (deru mobil) yang berbeda, ketika mobil mendekat dan menjauh. Rata-rata mobil balap mengeluarkan bunyi 800 Hz. Jika kecepatan bunyi di udara 340 m.s-1 dan kecepatan mobil 20 m.s-1, maka frekuensi yang di dengar saat mobil NEWTON’S SECOND LAW OF MOTION Solved problems in Newton’s laws of motion – Newton’s second law of motion 1. A 1 kg object accelerated at a constant 5 m/s2. Estimate the net force needed to accelerate the object. Known : Mass (m) = 1 kg Acceleration (a) = 5 m/s2 Wanted : net force (∑F) Solution : We useNewton’s
HOOKE’S LAW
1. A graph of force (F) versus elongation (x) shown in the figure below. Find the spring constant! Solution Hooke’s law formula : k = F / x F = force (Newton) k = spring constant (Newton/meter) x = the change in length (meter) Spring constant : k = 10 / 0.02 = 20 /CARNOT CYCLE
6. An Carnot engine absorbs heat at high temperature 800 Kelvin and efficiency of the Carnot engine is 50%. What is the high temperature to increase efficiency to 80% if the low temperature kept constant. Known : If high temperature (TH) = 800 K , efficiency (e) = 50% = 0.5. Wanted : High temperature (TH) if efficiency (e) = 80% = 0.8. Solution:
WORK-MECHANICAL ENERGY PRINCIPLE Work-mechanical energy principle – problems and solutions. 1. A 2-kg box decelerated from 10 m/s to rest. The coefficient of kinetic friction is 0.2. Acceleration due to gravity is 10 m/s2. AVERAGE SPEED AND AVERAGE VELOCITY Solved Problems in Linear Motion – Average speed and average velocity 1. A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second. Determine average speed and average velocity. Solution Distance = 100 meters + 50 meters = 150 meters Displacement KINETIC THEORY OF GASES Kinetic theory of gases – problems and solutions. 1. Ideal gases in a closed container initially have volume V and pressure P. If the final pressure is 4P and the volume is kept constant, what is the ratio of the initial kinetic energy with the final kinetic energy. Known : Initial pressure (P1) = P. DISTANCE AND DISPLACEMENT Solved Problems in Linear Motion – Distance and displacement 1. A car travels along a straight road 100 m east then 50 m west. Find distance and displacement of the car. Solution Distance is 100 meters + 50 meters = 150 meters Displacement is 100 meters – 50 meters = 50 meters, to the east.LINEAR EXPANSION
1. A steel is 40 cm long at 20 oC. The coefficient of linear expansion for steel is 12 x 10-6 (Co)-1. The increase in length and the final length when it is at 70 oC will be Known : The change in temperature (ΔT) = 70oC – 20oC = 50oC The original length (L1) CAPACITORS IN SERIES AND PARALLEL 1. Three capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 4 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination. Known : Capacitor C1 = 2 μF Capacitor C2 = 4 μF Capacitor C3 = 4 μF Wanted PEMBAHASAN SOAL GELOMBANG BUNYI Pembahasan soal gelombang bunyi. 1. Soal UN 2008/2009 P04 No.24. Seorang penonton pada lomba balap mobil mendengar bunyi (deru mobil) yang berbeda, ketika mobil mendekat dan menjauh. Rata-rata mobil balap mengeluarkan bunyi 800 Hz. Jika kecepatan bunyi di udara 340 m.s-1 dan kecepatan mobil 20 m.s-1, maka frekuensi yang di dengar saat mobilLINEAR MOTION
Linear motion – problems and solutions. 1. Graph of velocity (v) vs. time (t) shown in the figure below. What is the deceleration according to the graph. Solution. A-B = motion at constant acceleration, B-C = motion at constant velocity, C-D = motion at constant deceleration.LINEAR EXPANSION
1. A steel is 40 cm long at 20 oC. The coefficient of linear expansion for steel is 12 x 10-6 (Co)-1. The increase in length and the final length when it is at 70 oC will be Known : The change in temperature (ΔT) = 70oC – 20oC = 50oC The original length (L1) POWER – PROBLEMS AND SOLUTIONS Power – problems and solutions. 1. A 50-kg person runs up the stairs 10 meters high in 2 minutes. Acceleration due to gravity (g) is 10 m/s2. Determine the power. W = work, F = force, w = weight, d = displacement, h = height, m = mass, g = acceleration due to gravity. WOHM’S LAW
Ohm’s law : V = I R or R = V / I. R1 = V / I = 1.50 / 0.08 = 18.75 Ohm. R2 = V / I = 2.80 / 1.50 = 2.87 Ohm. R3 = V / I = 3.99 / 2.10 = 1.9 Ohm. Based on the table, the relationship between V, I and R is the value of R inversely proportional to V and I. That means thelarger the V
COULOMB’S LAW
Coulomb’s law – problems and solutions. 1. Two point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude of the electric force. The constant k = 8.988 x 109 Nm2C. Advertisement. Advertisement. −2 = 9 x 109 Nm2C−2.INTENSITY OF SOUND
Solution : The addition of the sound intensity of 10 W/m2 = 10-3 W/cm2 is equivalent to the addition of the sound intensity level of 10 dB. The addition of sound intensity of 102 W/m2 = 10-2 W/cm2 is equivalent to the addition of sound intensity level of 20 dB. And so on. If the intensity level is increased by 10 dB, the intensity increases by DENSITY – PROBLEMS AND SOLUTIONS Mass (m) = 300 gram. Volume (V) = volume of spilled water = 20 cm3. Wanted : density. Solution : ρ = m / V = 300 gram / 20 cm3 = 15 gram/cm3. Read : Heat transfer conduction – problems and solutions. 7. Mass of object is 316 gram, placed in a container as shown in figure. What is the density of the object. AVERAGE SPEED AND AVERAGE VELOCITY Solved Problems in Linear Motion – Average speed and average velocity 1. A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second. Determine average speed and average velocity. Solution Distance = 100 meters + 50 meters = 150 meters DisplacementBUOYANT FORCE
Wanted : Density of gold. Solution : Weight of gold in air : w = m g = ρb V g —– Equation 1. V is volume of gold in kerosene. Buoyant force (FB) equal to the weight of gold in air (w) minus weight of gold in kerosene (w’) : w – w’ = FA. w – w’ = ρf V g —– Equation 2. V is the volume of gold in kerosene. CAPACITORS IN SERIES AND PARALLEL 1. Three capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 4 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination. Known : Capacitor C1 = 2 μF Capacitor C2 = 4 μF Capacitor C3 = 4 μF Wanted* Fisika X
* Fisika XI
* Fisika XII
* UN Fisika
* Artikel Fisika
* Fisika SMP
* Fisika SMA
* Pembahasan Soal Fisika SMP * Pembahasan Soal Fisika SMA* UN Fisika SMP
* UN Fisika SMA
* Soal Fisika SMP
* Soal Fisika SMA
* Percobaan Fisika SMA* OSN Fisika SMP
* Fisika Dasar
Main MenuFisika XFisika XIFisika XIIUN FisikaArtikel FisikaFisika SMPFisika SMAPembahasan Soal Fisika SMPPembahasan Soal Fisika SMAUN Fisika SMPUN Fisika SMASoal Fisika SMPSoal Fisika SMAPercobaan Fisika SMAOSN Fisika SMPFisika Dasar GURUMUDA.NET - PELAJARAN FISIKA SEKOLAH MENENGAH PEMBAHASAN SOAL UN FISIKA SMA TAHUN 2018 (21-40) 21. Asas Black Air mendidih (100oC) sebanyak 250 ml dituangkan ke dalam panci berisi 400 ml air bersuhu 35oC. Setelah terjadi keseimbangan termal, maka suhu campuran adalah… (kalor jenis air 1,0 kal.gr-1.oC-1) Pembahasan Diketahui: Suhu air panas = 100... PEMBAHASAN SOAL UN FISIKA SMA TAHUN 2018 (1-20) 1. Pengukuran Pada saat melakukan praktikum pengukuran dengan menggunakan jangka sorong, seorang siswa mengukur dimensi balok tinggi, panjang dan lebar dengan hasil pengukuran sebagai berikut. Volume balok tersebut sesuai kaidah angka penting adalah… Pembahasan Tinggi balok: Skala utama =... PEMBAHASAN SOAL UN FISIKA SMA TAHUN 2017 (21-40) 21. Elastisitas Perhatikan tabel data hasil percobaan pegas berikut ini! Pegas diberi beban dan diukur pertambahan panjangnya. Dari pernyataan berikut : (1) Pertambahan panjang pegas sebanding dengan berat beban (2) Pertambahan panjang pegas berbanding terbalik denganberat beban (3)...
PEMBAHASAN SOAL UN FISIKA SMA TAHUN 2017 (1-20) 1. Kecepatan rata-rata Sebuah partikel bergerak ke arah utara dari titik awal (0,0) menuju titik P (2, 4) dalam waktu 3 sekon. Partikel kemudian membelok ke timur selama 2 sekon sampai di titik Q (10, 10). Besar kecepatan rata-rata... PEMBAHASAN SOAL UN FISIKA SMA TAHUN 2016 (21-40) 21. Sifat Gas Ideal Soal UN fisika SMA/MA 2015/2016 No.21 Perhatikan pernyataan di bawah ini! (1) Setiap partikel selalu bergerak dengan arah tertentu (2) Partikel gas tersebar merata ke seluruh ruangan (3) Ukuran partikel gas dapat diabaikan terhadap ukuran... PEMBAHASAN SOAL UN FISIKA SMA TAHUN 2016 (1-20) 1. Pengukuran fisika Soal UN fisika SMA/MA 2015/2016 No.1 Pengukuran diameter silinder baja dengan mikrometer sekrup adalah 4,47 mm. Hasil pengukuran yang benar ditunjukkan pada gambar berikut adalah… Pembahasan A. Skala utama = 4,5 mm Skala putar = 47... PEMBAHASAN SOAL UN FISIKA SMA TAHUN 2015 (21-40) 21. Alat optik mikroskop Soal UN fisika SMA/MA 2014/2015 No.21 Sebuah mikroskop mempunyai jarak fokus obyektif dan fokus okulernya masing-masing 0,9 cm dan 2,5 cm digunakan oleh orang bermata normal tanpa berakomodasi dan ternyata perbesarannya 90 kali. Berartijarak...
PEMBAHASAN SOAL UN FISIKA SMA TAHUN 2015 (1-20) 1. Pengukuran Fisika Soal UN fisika SMA/MA tahun 2014/2015 No.1 Pengukuran diameter kawat dengan mikrometer sekrup adalah 2,48 mm. Gambar yang sesuai dengan hasil pengukuran tersebut adalah… Pembahasan A. Diameter kawat = 4,….. B. Diameter kawat = 3,…..C....
CONTOH SOAL RESISTOR PARALEL 1. Diketahui R1 = 2 Ω dan R2 = 4 Ω tersusun paralel. Berapa nilai resistor pengganti ? (Ω = Ohm). Pembahasan 1/R = 1/R1 + 1/R2 = 1/2 + 1/4 = 2/4 + 1/4 = 3/4 R =... CONTOH SOAL RESISTOR SERI 1. Diketahui R1 = 2 Ω dan R2 = 4 Ω tersusun seri. Berapa nilai resistor pengganti ? (Ω = Ohm). Pembahasan R = R1 + R2 = 2 + 4 = 6 Ω. Nilai hambatan pengganti lebih besar... PAGES 1 OF 75:1 2 34 5
6 »
... Last »
KATEGORI ARTIKEL
* Artikel Fisika
* Fisika SMP
* Fisika SMA
* Pembahasan Soal Fisika SMP * Pembahasan Soal Fisika SMA* UN Fisika SMP
* UN Fisika SMA
* Soal Fisika SMP
* Soal Fisika SMA
* Percobaan Fisika SMA* OSN Fisika SMP
* Fisika Dasar
Tentang | Kontak
| Pustaka
| Privasi
| Ketentuan
Copyright © Gurumuda.Net All Rights Reserved.* Fisika X
* Fisika XI
* Fisika XII
* UN Fisika
* Artikel Fisika
* Fisika SMP
* Fisika SMA
* Pembahasan Soal Fisika SMP * Pembahasan Soal Fisika SMA* UN Fisika SMP
* UN Fisika SMA
* Soal Fisika SMP
* Soal Fisika SMA
* Percobaan Fisika SMA* OSN Fisika SMP
* Fisika Dasar
7.8k Shares
Share
Tweet
Details
Copyright © 2024 ArchiveBay.com. All rights reserved. Terms of Use | Privacy Policy | DMCA | 2021 | Feedback | Advertising | RSS 2.0