The distance from the Earth to the Sun is 92.96 million mi.
Answer:
(a) the deceleration of the player is -80.36 m/s²
(b) the time the collision last is 0.093 s
Explanation:
Given;
Initial velocity of the football player, u = 7.50 m/s
Final velocity of the football player, v = 0
distance traveled = compression of the pad, s = 0.35 m
Part (a) the deceleration of the player
v² = u² + 2as
0 = 7.5² + (2 x 0.35)a
0 = 56.25 + 0.7a
- 56.25 = 0.7a
a = -56.25 / 0.7
a = -80.36 m/s²
Part (b) the time the collision last
v = u + at
t = (v - u)/a
t = (0 - 7.5)/ -80.36
t = - 7.5 / -80.36
t = 0.093 s
<span>Work done on charge is W = Eqd = σ/(2ε₀) x q x d = {(8.00 x 10⁻¹²)/(2 x 8.854187 x 10⁻¹²)} x 3.00 x 10⁻⁶ x (0.650 - 0.250) = 5.42116402J. KE of sphere = 0.5mv² = 0.5 x 5.00 x 10⁻⁷v² = work done by E-field on charge during its fall = 5.42116402→ v = 4657 m/s.</span>
Answer:
2.9
Explanation:
average speed=distance/time taken
=2/5.8
=2.9
