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GED0081L
Physics Lab Formas
| Question | Answer |
|---|---|
| It is the reciprocal of period T. | frequency |
| It is the reciprocal of frequency f. | period |
| Centripetal force is _____ proportional to mass of the rotating object. | directly |
| Centripetal force is _____ proportional to centripetal acceleration. | directly |
| Centripetal force is _____ proportional to speed of the rotating body. | directly |
| Centripetal force is _____ proportional to radius of the circular path (curvature) of the rotating body. | directly |
| Centripetal force is _____ proportional to frequency of rotation of the rotating body. | directly |
| What is the relationship between the counterweight and centripetal force? | equal |
| Calculate the frequency of the rotation object which completes 250 rotations in 10 seconds. | 25 Hertz |
| If a rotating object has a frequency of 1 KHz. Calculate its period. | 1 milisecond |
| t is aimed that a 500-N balikbayan box be moved from rest along a flat surface by applying a pull. The box started to slide as it was pulled with an applied force of 150N. The box kept on sliding with a constant pull. What is the static frictional force? | 150N |
| It is aimed that a 500-N balikbayan box be moved from rest along a flat surface by applying a pull. The box started to slide as it was pulled with an applied force of 150N. The box kept on sliding with a constant pull. What is the kinetic frictional force | less than 150N |
| t is aimed that a 500-N balikbayan box be moved from rest along a flat surface by applying a pull. The box started to slide as it was pulled with an applied force of 150N. The box kept on sliding with a constant pull. What is the coefficient of static fri | 0.3 |
| 500-N balikbayan box... Applied force of 150N... Assuming the platform has a coefficient of static friction of 0.95, the coefficient of kinetic friction is: | less than 0.95 |
| 500-N balikbayan box... Applied force of 150N... If the platform is smooth, free from friction, what is the coefficient of static friction? | 0 |
| The coefficient of static friction is always _____ the coefficient of kinetic friction. | higher than |
| The static frictional force is always _____ the kinetic frictional force. | higher than |
| It is a force that prevents an object from starting to move. | Static friction |
| It is a frictional force acting on a moving object. | Kinetic friction |
| The frictional force can be computed by multiplying the _____ force by the coefficient of friction. | normal |
| It is referred to as energy of motion. | kinetic energy |
| It is the product of the moving object's mass 𝑚 and speed 𝑣. | kinetic energy |
| It is the product of the moving object's mass 𝑚 and its height h. | potential energy |
| This law states that energy can neither be created nor destroyed, only converted from one form to another. | Law of Conservation |
| The potential energy of a solid object at rest, when acted upon by a force that causes it to move, will be converted to _____ energy. | kinetic |
| Initially the potential energy of a stationary object, suspended by a string, is 10 kilo Joules. When the string breaks, the object falls down. What happens to its potential energy while the object is falling down? | decreasing |
| Initially the potential energy of a stationary object, suspended by a string, is 10 kilo Joules. When the string breaks, the object falls down. What happens to its kinetic energy while the object is falling down? | increasing |
| Initially the potential energy of a stationary object, suspended by a string, is 10 kilo Joules. When the string breaks, the object falls down. What happens to its total energy while the object is falling down? | no change |
| The potential energy of an stationary object is 10 kilo Joules. If the object is suspended by a string, 10 meters high, what is its potential energy? | 102 Joules |
| Initially the potential energy of an stationary object, suspended by a string, is 10 kilo Joules. When the string breaks, the objects falls down. At a point when its potential energy is down to 2 kilo joules, how much is its kinetic energy? | 8 kilo Joules |
| It is the product of mass and velocity. | Momentum |
| This is a type of collision where both momentum and kinetic energy are conserved | elastic collision |
| This is a type of collision where only the momentum is conserved but the kinetic energy is not. | Inelastic collision |
| The figure below shows what type of collision? figure: Billiards | elastic collision |
| This figure below shows what type of collision? figure: Before Collision and After Collision of 2 Cars where 1 car loses speed. | Inelastic collision |
| This figure below shows what type of collision? figure: Newton's Cradle | Elastic collision |
| This figure below shows what type of collision? figure: Bowling ball Strike | Elastic collision |
| This figure below shows what type of collision? figure: American football man getting tackled | Inelastic collision |
| This figure below shows what type of collision? figure: Man playing golf | elastic collision |
| A 60-kg body is moving with a velocity of 20 m/s. How much is it momentum? | 1200 kg.m/s |
| For the system to be at equilibrium, it should meet one of the two conditions which is: Net force is equal to : | 0 |
| For the system to be at equilibrium, it should meet one of the two conditions which is: Net torque is equal to : | 0 |
| It is the tendency of an object to rotate about an axis due to a force. | Torque |
| Find the torque of the system shown below: r = 2m F = 10N (downwards) | T = 20Nm |
| The system below will rotate in r = 2m F = 10N (downwards) | CW : Clock Wise |
| The system below will rotate in: r = 2m F = 10N (downwards) F = 20N (upwards) | CCW : Counter Clock Wise |
| Find the torque of the system shown below: r = 2m F = 10N (downwards) F = 20N (upwards) | 20 Nm |
| What is the net torque of the system below: r1 = 0.5m F1 = 10N (downwards) r2 = 2m F2 = 20N (upwards) F2 = 10N(downwards) | 15 Nm |
| The system below will rotate in: left: r = 3m r1 = 1m F1 = 10N r2 = 2m F2 = 10N r3 = 3m F3 = 10 N right: r = 3m F = 20m | CCW |
| The system below will rotate in: left: r = 1m F = 10m right: r = 2m F = 5N | is at equilibrium, no rotation |
Created by:
AdoKwatro