1. a 1100 kg autoobil travels at 80 km/h partA what is the kinect energy?K=___J

Question

1. a 1100 kg autoobil travels at 80 km/h

partA what is the kinect energy?K=___J

part B what net work would be required to bring it to a stop?W=__J

2.A 2.2 kg mud ball drops from rest at a height of 13 m .

part b If the impact between the ball and the ground lasts 0.50 s , what is the average net force exerted by the ball on the ground?

3.Find the magnitude of the linear momentum of 7.5 kg bowling ball traveling at 13 m/s.=___kg*m/s (two significant firgures.)

part b fid the magnitude of the linearbmomentum of a 1200 kg automobile traveling at 70 km/hp=__-kg*m/s

Explanation / Answer

3)

Potential energy PE (height * weigh) = kinetic energy KE (1/2)mV^2 = Work = average force times distance traveled in 0.5 sec.

Those are some basics; we will use what we need.

We know velocity at impact = V = sqrt(2gh) = sqrt(2*9.8*13) = 15.96 m/s
We know that F =ma and that distance D = (1/2)at^2 = 0.5a(0.5)^2 = 0.125a with a being the average acceleration during impact.
We know V = at = 15.96 m/s = a*0.5sec
a = 31.92 m/s^2
F= ma = 2.2 kg*31.92 m/s^2 = 70.224 N

4)

P=m*v
a)

P=7.5*13 = 97.5 kg- m/s

b)

70 km/h = 19.44 meters / second

P= 1200*19.44 = 23328 kg-m/s

2)

x = Vo t cos

t = x/(Vo cos )

y = yo + Vo t sin + ½ g t²

y = yo + Vo (x/(Vo cos )) sin + ½ g (x/(Vo cos ))²

y = yo + x tan + ½ g (x/(Vo cos ))²

3.05 = 2.05 + 6.02 tan 34° + ½ (-9.81) (6.02/(Vo cos 34°))²

3.05 = 6.11 - 258.64 / Vo^2

Vo = 9.1936 m/s

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.