Answer:
the faster an object moves the more kinetic it has. the more mass an object has, the more kinetic energy it has.
The distance traveled while accelerating from rest is
D = 1/2 a t² .
For this problem, we shall totally ignore air resistance.
We do so completely without any reservation or guilt,
because we know that there is no air on the moon.
D = (1/2) · (1.6 m/s²) · (9 sec)²
= (0.8 m/s²) · (81 s²)
= (0.8 · 81) m
= 64.8 meters .
(That's about 213 feet ! The astronaut must have dropped the feather
from his spacecraft while he was aloft ... either just before touchdown
or just after liftoff.)
Answer:
Jane's measurements are precise.
Explanation: Jane's values are close to each other, so they are precise.
Answer:
a) Average Speed = 190yards/26.6s = 7.14 yards/s
b) Average Velocity = 10/26.6 = 0.38 yard/s
Complete Question;
A football player runs from his own goal line to the opposing team's goal line, returning to his ten-yard line, all in 26.6 s. Calculate his average speed and the magnitude of his average velocity. (Assume a 100 yard football field; note that the player's X-yard line is X yards from his own goal line. Enter your answers in yards/s.)
Explanation:
a) Speed = distance/time
time t = 26.6s
distance;
d1 = 100 yards (own goal line to the opposing team's goal line)
d2 = 90 yards (returning to his ten-yard line)
d = d1 + d2
d = 100 + 90 = 190 yards
Speed = 190yards/26.6s = 7.14 yards/s
b) velocity = displacement/time
time = 26.6s
displacement = 100-90 yards = 10yards
Velocity = 10/26.6 = 0.38 yard/s
