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Physics

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A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 35 m/s in 0.050 s. If the mass of the paper is 0.0045 kg, what force does the boxer exert on it?
____ N
Assuming no air resistance, we calculate the average acceleration of the paper...
a = (vf  vi) / t =
(35m/s  0m/s) / 0.050 s = 700 m/s^2
To get the force, calculate....
F=ma
F = (0.0045 kg)* (700 m/s^2) = 3.15 N
thanks can you help me with
A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 35 m/s in 0.050 s. If the mass of the paper is 0.0045 kg, what force does the boxer exert on it?
____ N
Newton's first law tells us that F = m*a
So, to calculate the force in newtons [N] I need to know both the mass and acceleration of the object (the paper) and then to multiply these two values together.
The mass of the paper we know, m = 0.0045 kg
The acceleration (a) we need to calculate. Acceleration is defined as the change in velocity per some time interval. So, we go from an initial velocity (vi) of 0 m/s (at rest) to a final velocity (vf) of 35 m/s. I takes the paper 0.050 seconds to go from this initial to final velocity. The change in velocity, sometimes written delta velocity, is calculated by taking the final velocity (vf) and subtracting the initial velocity (vi).
So acceleration (a) is equal to...
a = (vf  vi) / t
where t is the time
a = (35 m/s  0 m/s) / 0.050 s =
700 m/s^2
Now we have mass and acceleration, simply multiply them together to get the force...
F = m*a = 0.0045 kg * 700 m/s^2 =
3.15 N
Does that explanation help? Did you have a another similar problem but with different values for mass and velocities?
Consider an airplane
that normally has an airspeed of 120 km/h in a 75 km/h crosswind blowing from
west to east. Calculate its ground speed when its nose is pointed north in the
crosswind.
____ km/h
Ok, we'll assume that the crosswind completely pushes the plane 75 km/h off course.
To find the answer, we need to do vector addition.
We need to calculate the resultant vector found by adding the normal speed with the speed of the crosswind.
Notice that the three vectors form a right triangle, with the resultant vector acting as the hypotenuse . We'll make use of the Pythagorean Theorem to solve for the magnitude of the resultant speed...
c^2 = a^2 + b^2
c = SQRT ( 120^2 + 75^2)
= SQRT (14400 + 5625)
= SQRT (20025)
= 141.51 km/hr
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