A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 20m/s in 0.044 S

Complete Question:

A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 20m/s in 0.044 Seconds.

a. what acceleration is imparted to the paper?

b. if the mass of the paper is 0.003 kg, what force does the boxer exert on it?

c. how much force does the paper exert on the boxer?

a. Acceleration = 454.55m/s².

b. Force = 1.36N

c. Force = -1.36N

Step-by-step explanation:

Given the following data;

Final velocity = 20m/s

Time = 0.044 seconds.

Since the paper is at rest, we know that it's initial velocity is equal to 0m/s.

a. To find acceleration;

Mathematically, acceleration is given by the equation;

Substituting into the equation, we have;

Acceleration = 454.55m/s²

b. To find the force exerted by the boxer;

Mass = 0.003kg

Force = mass * acceleration

Force = 0.003 * 454.55

Force = 1.36N

c. To find the force exerted by the paper;

According to Newton's third law of motion which states that, for every action there is an equal and opposite reaction. Therefore, the paper exerts an equal but opposite force on the boxer.

Force = -1.36N

the function makes a strait line horizontally with no x intercept but a y intercept at 9.

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

d