A football punter accelerates a football from rest to a speed of 15 m/s during t
ID: 1415373 • Letter: A
Question
A football punter accelerates a football from rest to a speed of 15 m/s during the time in which his toe is in contact with the ball (about 0.16 s). If the football has a mass of 0.49 kg, what average force does the punter exert on the ball? N A boat moves through the water with two forces acting on it. One is a 1,725-N forward push by the water on the propeller, and the other is a 1,300-N resistive force due to the water around the bow. What is the acceleration of the 1,500-kg boat? m/s^2 If it starts from rest, how far will the boat move in 25.0 s? m What will its velocity be at the end of that time? m/s The force exerted by the wind on the sails of a sailboat is 330 N north. The water exerts a force of 190 N east. If the boat (including its crew) has a mass of 260 kg, what are the magnitude and direction of its acceleration? Magnitude m/s^2 direction A 130.6-N bird feeder is supported by three cables as shown In the figure below. Find the tension in each cable. Left cable right cable bottom cableExplanation / Answer
Part 1)
F=ma
a=(V-Vo)/t therefore
F=m [(V-Vo)/t]
Vo=0m
V= 15m/s
t=.16s
m= .49kg
S0,
F=0.49x(15-0)/0.16
=45.9375 N
Ans- average force is 45.9375 N
Part 2)
The net force on the boat is 425N.
The acceleration is found from Newton's Second Law
F = ma
425 N = ( 1500 kg ) a
a = .283 m/s²
The distance traveled is found from
x = 1/2 a t²
x = 1/2 ( .283 m/s² ) ( 25.0 s )²
x = 88.54 m
The final velocity is
v = at
v = ( .283 m/s² ) ( 25 s )
v = 7.075 m/s
ans a) acceleration is .283 m/s²
b)Boat will move 88.54 m in 25 sec
c) Its velocity at the end of that time will be 7.075 m/s
Part 3)
magnitude:
The two forces are at right angles so Pythagorean theorem can be applied.
F^2 = 330^2 + 190^2
F = 380.78N
F = ma
380.78 = 260a
a = 1.464 m/s^2
Direction:
It goes north east a bit.
tan alpha = 330/260
alpha = 51.76degrees
ans- Magnitude of acceleration=a = 1.464 m/s^2
Direction= alpha = 51.76degrees
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