(a) If you ride in an elevator moving upward with constant speed, is your apparent weight greater than, less than, or equalto your actual weight, mg? (b) Which of the following is the best explanation for your prediction?I. The elevator is moving upward. This will cause a person to press down harder on the floor, giving an apparent weightgreater than mg.II. The elevator moves with zero acceleration, and therefore the apparent weight is equal to mg.III. Because the elevator is moving upward it is partially lifting the person, resulting in an apparent weight less than mg

1. Solve the following problems. a) What is the mass of a book that weighs 3 N at a location where g = 9.8 m/s2? b) At the same location, what is the weight of a dog whose mass is 12 kg? 2. At the surface of Mars, the acceleration due to gravity is g = 3.7 m/s2. A watermelon weighs 45 N at the surface of the earth. What are its mass and weight on the surface of Mars? 3. A crate with mass 4 kg and initially at rest on a frictionless horizontal plane is acted upon by a horizontal force of 10 N. Calculate : a) What acceleration is produced? b) How far does the crate travel in 10 seconds? c). What is its speed at the end of 10 seconds?

A 12 kg meteor experiences an acceleration of 7.2 m/s*2, when falling towards the earth. a) How high above the earth's surface is the meteor? b) What force will a 30 kg meteor experience at the same altitude?

A parachutist  wants to design a parachute, such that a person weighing 150 lbs(667N) descends at a rate of 5.00 mph(2.23 m/s). To calculate the drag coefficient in kg/m, is the way to conceptualize the solution: 1. At terminal velocity the drag force equals the force of gravity. Since the net forces are zero, there is no acceleration and the velocity is constant.   2. The drag constant b is not 1. It comes from the terminal velocity equation: V_term  = square root(mg/b) If you know the weight mg and the terminal velocity, then you can solve for b.   Please help me to derive the solution by helping me to understand better.  I understand point 1.  But I don't understand clearly point 2?

Figaro the cat (5.0 kg) is hanging on the tablecloth, pulling Cleo’sfishbowl (11 kg) toward the edge of the table. The coefficient ofkinetic friction between the tablecloth (ignore its mass) under thefishbowl and the table is 0.44. (Hint: treat the table cloth as a rope– this is a modified atwood’s machine)a. What is the acceleration of Figaro and the fishbowl?b. Using the acceleration in part (a), find the time for Figaroto pull Cleo off the table if the fish bowl is .90 m away fromthe edge of the table.

A 38.0 kg box hangs from a rope. What is the tension (in N) in the rope if the box is at rest? Assume that the local acceleration due to gravity is -9.80 m/s2.  What is the tension (in N) in the rope if the box moves up at a steady 5.0 m/s? What is the tension (in N) in the rope if the box has an initial velocity of 5.0 m/s and is speeding up at 5.0 m/s2?  What is the tension (in N) in the rope if the box has an initial velocity of 5.0 m/s and is slowing down at 5.0 m/s2?

Imagine holding a 25kg dumbbell in your hand. a) What force must you exert on the dumbbell to hold it still? b) What force must you exert on the dumbbell to keep it moving upward at constant velocity?

Two masses of weight 30.0 NN are suspended at opposite ends of a rope that passes over a light, frictionless pulley. The pulley is attached to a chain that goes to the ceiling. What is the tension in the rope? What is the tension in the chain?

What is the answer and draw the FBD. In an airplane crash a woman is holding an 8.18 kg, 18 pound, baby. In the crash the woman experiences a horizontal de-acceleration of 88.2 m/s2. How many g's is this de-acceleration? How much force must the woman exert to hold the baby in place?

I have a question for elevator problems: Newton's law states that there is always an equal and opposite force. But for elevator problems, if it accelerates upwards... we say that the normal force is greater than the gravitational force. So is Newton's third law still applied here and if so how? Because technically the opposite force is not equal in magnitude for it to be able to accelerate