Answer:
0.0337 J or 33.7 mJ
Explanation:
PEspring = ½ kx = 0.5 A 55.0 N/m A (0.0350 m) = 0.0336875 J = 2 2
Answer:
E=147898.01J
Explanation:
A 5.0-kg cannonball is fired over level ground with a velocity of 2.00 ⨯ 102 m/s at an angle of 25° above the horizontal. Just before it hits the ground its speed is 150 m/s. Over the entire trip, find the change in the thermal energy of the cannonball and air
firstly , we look for the time of flight it takes to make the projectile path
T=2Usin∅/g
take g= 9.81m/s
T=2*200sin25/(9.81)
T=17.23Secs
energy is force *distance
E=f*d
f=m*g
f=5*9.81
f=49.05N
s=distance
s=(v+u)T/2
s=(150+200)17.23/2
s=3015.25m
49.05N*3015.25m
E=147898.01J
Answer:
If you want help, try showing that diagram and the following statements.