Answer:
W = 1.545E6 J total work
P = W / t = 1.545E6 J / 3 sec = 5.15E5 J/sec = 515,000 J/sec (Watts)
Using definition of power
Answer:
s = 23.72 m
v = 21.56 m/s²
Explanation:
given
time to reach the ground (t) = 2.2 second
we know that
a) s = u t + 0.5 g t²
u = 0 m/s
g = 9.8 m/s²
s = 0 + 0.5 × 9.8 × 2.2²
s = 23.72 m
b) impact velocity
v = √(2gh)
v = √(2× 9.8 × 23.72)
v = √464.912
v = 21.56 m/s²
Answer:
d) 289.31 m
Explanation:
Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m
Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .
Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d
To equate
357.18 m -144.54 m = .735 m d
d = 289.31 m .
