If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45.
Incomplete question.The complete one is here
A runner taking part in the 200m dash must run around the end of a track that has a circular arc with a radius curvature of 30m. The runner starts the race at a constant speed. If she completes the 200m dash in 23s and runs at constant speed throughout the race, what is her centripetal acceleration as she runs the curved portion of the track?
Answer:
Explanation:
Given data
Required
Centripetal acceleration
Solution
According to the motions equation the velocity given by:
The centripetal acceleration is given by:
Answer: the answer would be C.
Explanation: hope this helps sorry if I got it wrong.
"2 km/hr/s" means that in each second, its engines can increase its speed by 2 km/hr.
If it keeps doing that for 30 seconds, its speed has increased by 60 km/hr.
On top of the initial speed of 20 km/hr, that's 80 km/hr at the end of the 30 seconds.
This whole discussion is of <em>speed</em>, not velocity. Surely, in high school physics,
you've learned the difference by now. There's no information in the question that
says anything about the train's <em>direction</em>, and it was wrong to mention velocity in
the question. This whole thing could have been taking place on a curved section
of track. If that were the case, it would have taken a team of ace engineers, cranking
their Curtas, to describe what was happening to the velocity. Better to just stick with
speed.
Answer:
V = S / d = 60 m / 1.5 sec = 40 m/s
Note 40 m/s = 40 * 3.28 = 131 ft/sec
131 ft/sec = 131 ft/sec / (88 ft/sec / 60 mph) = 89 mi/hr
