Running on sand requires 1.6 times more energy spent than running on hard surface, so the force applied by our foot on sand is less.
First we write the corresponding kinematics equations:
a = -g
v = -g * t + vo
y = -g * ((t ^ 2) / 2) + vo * t + yo
Substituting the values:
y = - (9.81) * (((0.50) ^ 2) / 2) + (19) * (0.50) + (0) = 8.27m
answer:
the displacement at the time of 0.50s is 8.27m
When the comet is closest to the Sun,
it has its maximum kinetic energy
and minimum gravitational potential energy. When the comet is far away from the Sun, it has maximum gravitational potential energy and minimal kinetic energy. It's faster when it's close because the Sun's gravity is pulling the comet closer. The opposite for when it gets farther away
Answer:
20 m/s
30 m/s
Explanation:
Given:
v₀ = -10 m/s
a = -9.8 m/s²
When t = 1 s:
v = v₀ + at
v = (-10 m/s) + (-9.8 m/s²) (1 s)
v = -19.8 m/s
When t = 2 s:
v = v₀ + at
v = (-10 m/s) + (-9.8 m/s²) (2 s)
v = -29.6 m/s
Rounded to one significant figures, the speed of the ball at 1 s and 2 s is 20 m/s and 30 m/s, respectively.
