# Davinder

### what is the increasing and decreasing, write interval notation?

2 AnswersMathematics2 months ago### A racquet ball with mass m = 0.225 kg is moving toward the wall at v = 15.5 m/s and at an angle of θ = 29?

A racquet ball with mass m = 0.225 kg is moving toward the wall at v = 15.5 m/s and at an angle of θ = 29° with respect to the horizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.079 s.

1)What is the magnitude of the initial momentum of the racquet ball?kg-m/s

2)What is the magnitude of the change in momentum of the racquet ball?kg-m/s

3)What is the magnitude of the average force the wall exerts on the racquet ball?N

4)Now the racquet ball is moving straight toward the wall at a velocity of vi = 15.5 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -10.5 m/s. The ball exerts the same average force on the ball as before.What is the magnitude of the change in momentum of the racquet ball?kg-m/s

5)What is the time the ball is in contact with the wall?s 6)What is the change in kinetic energy of the racquet ball?J

3 AnswersPhysics2 months ago### A racquet ball with mass m = 0.225 kg is moving toward the wall at v = 15.5 m/s and at an angle of θ = 29°?

A racquet ball with mass m = 0.225 kg is moving toward the wall at v = 15.5 m/s and at an angle of θ = 29° with respect to the horizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.079 s.

1)What is the magnitude of the initial momentum of the racquet ball?kg-m/s

2)What is the magnitude of the change in momentum of the racquet ball?kg-m/s

3)What is the magnitude of the average force the wall exerts on the racquet ball?N

4)Now the racquet ball is moving straight toward the wall at a velocity of vi = 15.5 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -10.5 m/s. The ball exerts the same average force on the ball as before.What is the magnitude of the change in momentum of the racquet ball?kg-m/s

5)What is the time the ball is in contact with the wall?s 6)What is the change in kinetic energy of the racquet ball?J

1 AnswerPhysics2 months ago### A bumper car with mass m1 = 102 kg is moving to the right with a velocity of v1 = 4.9 m/s. A second bumper ?

A bumper car with mass m1 = 102 kg is moving to the right with a velocity of v1 = 4.9 m/s. A second bumper car with mass m2 = 90 kg is moving to the left with a velocity of v2 = -3.3 m/s. The two cars have an elastic collision. Assume the surface is frictionless.

3)What is the final velocity of car 1 in the center-of-mass reference frame?m/s

4)What is the final velocity of car 1 in the ground (original) reference frame?m/s

5)In a new (inelastic) collision, the same two bumper cars with the same initial velocities now latch together as they collide.

What is the final speed of the two bumper cars after the collision?

2 AnswersPhysics2 months ago### A white billiard ball with mass mw = 1.3 kg is moving directly to the right with a speed of v = 3.44 m/s ?

A white billiard ball with mass mw = 1.3 kg is moving directly to the right with a speed of v = 3.44 m/s and collides elastically with a black billiard ball with the same mass mb = 1.3 kg that is initially at rest. The two collide elastically and the white ball ends up moving at an angle above the horizontal of θw = 61° and the black ball ends up moving at an angle below the horizontal of θb = 29°.

1)What is the final speed of the white ball?m/s

2)What is the final speed of the black ball?m/s

3)What is the magnitude of the final total momentum of the system?kg-m/s

4)What is the final total energy of the system?J

1 AnswerPhysics2 months ago### An object with total mass mtotal = 14.5 kg is sitting at rest when it explodes into three pieces. ?

An object with total mass mtotal = 14.5 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 kg moves up and to the left at an angle of θ1 = 19° above the –x axis with a speed of v1 = 26.6 m/s. A second piece with mass m2 = 5.4 kg moves down and to the right an angle of θ2 = 24° to the right of the -y axis at a speed of v2 = 23.2 m/s.

1)What is the magnitude of the final momentum of the system (all three pieces)?

kg-m/s

2)What is the mass of the third piece?

kg

3)What is the x-component of the velocity of the third piece?

m/s

4)What is the y-component of the velocity of the third piece?

m/s

5)What is the magnitude of the velocity of the center of mass of the pieces after the collision?

m/s

6)Calculate the increase in kinetic energy of the pieces during the explosion.

J

1 AnswerPhysics2 months ago### An object with total mass mtotal = 14.5 kg is sitting at rest when it explodes into three pieces. ?

An object with total mass mtotal = 14.5 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 kg moves up and to the left at an angle of θ1 = 19° above the –x axis with a speed of v1 = 26.6 m/s. A second piece with mass m2 = 5.4 kg moves down and to the right an angle of θ2 = 24° to the right of the -y axis at a speed of v2 = 23.2 m/s.

1)What is the magnitude of the final momentum of the system (all three pieces)?

kg-m/s

2)What is the mass of the third piece?

kg

3)What is the x-component of the velocity of the third piece?

m/s

4)What is the y-component of the velocity of the third piece?

m/s

5)What is the magnitude of the velocity of the center of mass of the pieces after the collision?

m/s

6)Calculate the increase in kinetic energy of the pieces during the explosion.

J

1 AnswerPhysics2 months ago### A racquet ball with mass m = 0.225 kg is moving toward the wall at v = 15.5 m/s and at an angle of θ = 29?

A racquet ball with mass m = 0.225 kg is moving toward the wall at v = 15.5 m/s and at an angle of θ = 29° with respect to the horizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.079 s.

1)What is the magnitude of the initial momentum of the racquet ball?kg-m/s 2)What is the magnitude of the change in momentum of the racquet ball?kg-m/s 3)What is the magnitude of the average force the wall exerts on the racquet ball?N 4)Now the racquet ball is moving straight toward the wall at a velocity of vi = 15.5 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -10.5 m/s. The ball exerts the same average force on the ball as before.What is the magnitude of the change in momentum of the racquet ball?kg-m/s 5)What is the time the ball is in contact with the wall?s 6)What is the change in kinetic energy of the racquet ball?J

2 AnswersPhysics2 months ago### A white billiard ball with mass mw = 1.3 kg is moving directly to the right with a speed of v = 3.44 m/s and collides elastically with a bla?

A white billiard ball with mass mw = 1.3 kg is moving directly to the right with a speed of v = 3.44 m/s and collides elastically with a black billiard ball with the same mass mb = 1.3 kg that is initially at rest. The two collide elastically and the white ball ends up moving at an angle above the horizontal of θw = 61° and the black ball ends up moving at an angle below the horizontal of θb = 29°.

1)What is the final speed of the white ball?m/s 2)What is the final speed of the black ball?m/s 3)What is the magnitude of the final total momentum of the system?kg-m/s 4)What is the final total energy of the system?J

1 AnswerPhysics2 months ago### A bumper car with mass m1 = 102 kg is moving to the right with a velocity of v1 = 4.9 m/s. A second bump?

A bumper car with mass m1 = 102 kg is moving to the right with a velocity of v1 = 4.9 m/s. A second bumper car with mass m2 = 90 kg is moving to the left with a velocity of v2 = -3.3 m/s. The two cars have an elastic collision. Assume the surface is frictionless.

1)What is the velocity of the center of mass of the system?1.056m/s 2)What is the initial velocity of car 1 in the center-of-mass reference frame?3.84m/s 3)What is the final velocity of car 1 in the center-of-mass reference frame?m/s 4)What is the final velocity of car 1 in the ground (original) reference frame?m/s 5)What is the final velocity of car 2 in the ground (original) reference frame?m/s 6)In a new (inelastic) collision, the same two bumper cars with the same initial velocities now latch together as they collide.What is the final speed of the two bumper cars after the collision?m/s

2 AnswersPhysics2 months ago### A person with mass m1 = 66 kg stands at the left end of a uniform beam with mass m2 = 94 kg and a length L = 3 m.?

A person with mass m1 = 66 kg stands at the left end of a uniform beam with mass m2 = 94 kg and a length L = 3 m. Another person with mass m3 = 63 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 8 kg (assume that the medicine ball is at the far right end of the beam as well). Let the origin of our coordinate system be the left end of the original position of the beam as shown in the drawing. Assume there is no friction between the beam and floor.

1)What is the location of the center of mass of the system?m 2)The medicine ball is throw to the left end of the beam (and caught). What is the location of the center of mass now?m 3)What is the new x-position of the person at the left end of the beam? (How far did the beam move when the ball was throw from person to person?)m 4)To return the medicine ball to the other person, both people walk to the center of the beam. At what x-position do they end up?m

2 AnswersPhysics3 months ago### A man with mass m1 = 52 kg stands at the left end of a uniform boat with mass m2 = 171 kg and a length L = 3.5 m.?

DO NOT ROUND!!! IF MY TEXT DOESN'T LOOK GOOD, I THINK ITS YAHOO PROBLEM

A man with mass m1 = 52 kg stands at the left end of a uniform boat with mass m2 = 171 kg and a length L = 3.5 m. Let the origin of our coordinate system be the man’s original location as shown in the drawing. Assume there is no friction or drag between the boat and water.1)What is the location of the center of mass of the system?m 2)If the man now walks to the right edge of the boat, what is the location of the center of mass of the system?m 3)After walking to the right edge of the boat, how far has the man moved from his original location? (What is his new location?)m 4)After the man walks to the right edge of the boat, what is the new location the center of the boat?m 5)Now the man walks to the very center of the boat. At what location does the man end up?m

1 AnswerPhysics3 months ago### You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, ?

DO NOT ROUND!!!

You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the earth with enough speed to make it to the moon. Some information that will help during this problem:mearth = 5.9742 x 1024 kgrearth = 6.3781 x 106 mmmoon = 7.36 x 1022 kgrmoon = 1.7374 x 106 mdearth to moon = 3.844 x 108 m (center to center)G = 6.67428 x 10-11 N-m2/kg21)On your first attempt you leave the surface of the earth at v = 5534 m/s. How far from the center of the earth will you get?m 2)Since that is not far enough, you consult a friend who calculates (correctly) the minimum speed needed as vmin = 11068 m/s. If you leave the surface of the earth at this speed, how fast will you be moving at the surface of the moon? Hint carefully write out an expression for the potential and kinetic energy of the ship on the surface of earth, and on the surface of moon. Be sure to include the gravitational potential energy of the earth even when the ship is at the surface of the moon!m/s 3)Which of the following would change the minimum velocity needed to make it to the moon?the mass of the earththe radius of the earththe mass of the spaceship

2 AnswersPhysics3 months ago### A mass m = 12 kg is pulled along a horizontal floor with NO friction for a distance d =7.2 m.?

DO NOT ROUND!!!

A mass m = 12 kg is pulled along a horizontal floor with NO friction for a distance d =7.2 m. Then the mass is pulled up an incline that makes an angle θ = 31° with the horizontal and has a coefficient of kinetic friction μk = 0.49. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of θ = 31° (thus on the incline it is parallel to the surface) and has a tension T =62 N.1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)J 2)What is the speed of the block right before it begins to travel up the incline?m/s 3)What is the work done by friction after the block has traveled a distance x = 3.5 m up the incline? (Where x is measured along the incline.)J 4)What is the work done by gravity after the block has traveled a distance x = 3.5 m up the incline? (Where x is measured along the incline.)J 5)How far up the incline does the block travel before coming to rest? (Measured along the incline.)m 6)On the incline the net work done on the block is:positivenegativezero

1 AnswerPhysics3 months ago### A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass ?

A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass reaches the bottom of its path it is moving at a speed v = 2.1 m/s and the tension in the rope is T = 18.6 N.

1)How long is the rope?m 2)What is the mass?kg 3)If the maximum mass that can be used before the rope breaks is mmax = 1.51 kg, what is the maximum tension the rope can withstand? (Assuming that the mass is still released from the horizontal.)N 4)Now a peg is placed 4/5 of the way down the pendulum’s path so that when the mass falls to its vertical position it hits and wraps around the peg.How fast is the mass moving when it is at the same vertical height as the peg (directly to the right of the peg)?m/s 5)Return to the original mass. What is the tension in the string at the same vertical height as the peg (directly to the right of the peg)?N

1 AnswerPhysics3 months ago### a mass m = 4.5 kg hangs on the end of a massless rope L = 2.02 m long. The pendulum is held horizontal and released from rest.?

A mass m = 4.5 kg hangs on the end of a massless rope L = 2.02 m long. The pendulum is held horizontal and released from rest.

1)How fast is the mass moving at the bottom of its path?m/s 2)What is the magnitude of the tension in the string at the bottom of the path?N 3)If the maximum tension the string can take without breaking is Tmax = 389 N, what is the maximum mass that can be used? (Assuming that the mass is still released from the horizontal and swings down to its lowest point.)kg 4)Now a peg is placed 4/5 of the way down the pendulum’s path so that when the mass falls to its vertical position it hits and wraps around the peg. As it wraps around the peg and attains its maximum height it ends a distance of 3/5 L below its starting point (or 2/5 L from its lowest point).How fast is the mass moving at the top of its new path (directly above the peg)?m/s 5)Using the original mass of m = 4.5 kg, what is the magnitude of the tension in the string at the top of the new path (directly above the peg)?N

2 AnswersPhysics3 months ago### A mass m = 13 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5032 N/m. Th?

A mass m = 13 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5032 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.44. The mass leaves the spring at a speed v = 3.7 m/s.

1)How much work is done by the spring as it accelerates the mass?J 2)How far was the spring stretched from its unstreched length?m 3)The mass is measured to leave the rough spot with a final speed vf = 1.5 m/s.How much work is done by friction as the mass crosses the rough spot?J 4)What is the length of the rough spot?m 5)In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its unstretched length?m 6)In this new scenario, what would the coefficient of friction of the rough patch need to be changed to in order for the block to just barely make it through the rough patch? 7)Return to a scenario where the blcok makes it throgh the entire rough patch. If the rough patch is lengthened to a distance of three times longer, as the block slides through the entire distance of the rough patch, the magnitude of the work done by the force of friction is:the samethree times greaterthree times lessnine times greaternine times less

2 AnswersPhysics3 months ago### A block with mass m = 12 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4784 N/m after being?

A block with mass m = 12 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4784 N/m after being compressed a distance x1 = 0.491 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d = 2.9 m long. For this rough path, the coefficient of friction is μk = 0.44.

1)How much work is done by the spring as it accelerates the block?J 2)What is the speed of the block right after it leaves the spring?m/s 3)How much work is done by friction as the block crosses the rough spot?J 4)What is the speed of the block after it passes the rough spot?m/s 5)Instead, the spring is only compressed a distance x2 = 0.14 m before being released.How far into the rough path does the block slide before coming to rest?m 6)What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when released?m 7)If the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is:the samethree times greaterthree times lessnine times greaternine times less

1 AnswerPhysics3 months ago### A mass m1 = 5.5 kg rests on a frictionless table and connected by a massless string to another mass m2 = 4.1 kg.?

A mass m1 = 5.5 kg rests on a frictionless table and connected by a massless string to another mass m2 = 4.1 kg. A force of magnitude F = 26 N pulls m1 to the left a distance d = 0.83 m.

1)How much work is done by the force F on the two block system?J 2)How much work is done by the normal force on m1 and m2?J 3)What is the final speed of the two blocks?m/s 4)How much work is done by the tension (in-between the blocks) on block m2?J 5)What is the tension in the string?N 6)The net work done by all the forces acting on m1 is:positivezeronegative7)What is the NET work done on m1?

2 AnswersPhysics3 months ago