Here if we assume that there is no air friction on both balls then we can say
now the acceleration is given as
so here both the balls will have same acceleration irrespective of size and mass
so we can say that to find out the time of fall of ball we can use
now from above equation we can say that time taken to hit the ground will be same for both balls and it is irrespective of its mass and size
Answer:
987 joules, 3.01s
Explanation:
(A)
from the attached diagram
net force, Fnet, pulling the crate up the ramp is given by
Fnet = FcosФ - WsinФ - Fr
where FcosФ is the component of horizontal force 290N resolved parallel to the plane
WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane
Fr = constant opposing frictional force
Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65
Fnet = 240.421 - 109.602 - 65
Fnet = 65.82N
Work done on the crate up the ramp, W, is given by
W = Fnet × d (distance up the plane)
W = 65.819 × 15
W = 987.285 joules
W = 987 joules (to 3 significant Figures)
(B)
to calculate the time of travel up the ramp
we use the equation of motion
where s = distance up the plane, 15m
u = Initial velocity of the crate, which is 0 for a body that is initially at rest
a = acceleration up the plane, given by
where m = mass of the crate, 20 kg
now,
from,
15 = 0 + 1.645
15 = 1.645
t = 3.01s (to 3 sig fig)
Ans: Let d is the distance from height to our eyes.
<span>Applying the Pythagoras theorem, we get,</span>
<span>Check image: </span>
