A falling ball hits the ground at 10 m/s in a vacuum. What would happen to the velocity of the ball air resistance were taken into account. Discuss using the work-energy theorem. A) Regardless of the presence of air, the forces remain constant. By the work energy theorem, there will be no change in the net work done on the object falling and therefore no change change in kinetic energy.

B) Air resistance will be a force opposing the motion. The net force on the ball changes but the speed remains constant. According to the work energy theorem, a decrease in the net force causes a decrease in work.

C) Air resistance will be a force opposing the motion and will result in negative work done. By the work energy theorem, this will reduce the net work done on the object falling and therefore reduce the total change in kinetic energy.

D) Air resistance will be a force opposing the motion and will result in more work done to move the same distance. By the work energy theorem, this will increase the net work done on the object falling and therefore increase the total change in kinetic energy.

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

a 2,000 calorie based diet.