Answer:
(c) at point 2, the ball is at its highest height do its PE is max. Also at ms height, velocity is zero therefore KE is zero.
F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.
Answer:
22m/s
Explanation:
To find the velocity we employ the equation of free fall: v²=u²+2gh
where u is initial velocity, g is acceleration due to gravity h is the height, v is the velocity the moment it hits the ground, taking the direction towards gravity as positive.
Substituting for the values in the question we get:
v²=2×9.8m/s²×25m
v²=490m²/s²
v=22.14m/s which can be approximated to 22m/s
Pretty sure it’s A. Hope this helps.
Answer:c-The gravitational effect when spacecraft flies close to the asteriod
Explanation:
Gravitational effect on the spacecraft gives an estimate that how big is the asteroid by experiencing its gravitational pull.
The amount of extra thrust required to maintain the trajectory of the spacecraft during its motion hints at the scientist about the size of the asteroid.
Gravitational pull is directly proportional to the mass of object so greater the mass, greater will be the pull.
