The distance covered is 115 m
Explanation:
The motion of Ileana is a uniformly accelerated motion (constant acceleration), therefore we can use the following suvat equation:
where
s is the distance covered
u is the initiaal velocity
v is the final velocity
t is the time elapsed
In this problem, we have:
u = 4.20 m/s
v = 5.00 m/s
t = 25.0 s
Therefore, we can re-arrange the equation to find the distance covered:
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(a) 198 g
When the rock is submerged into the water, there are two forces acting on the rock:
- its weight, equal to (m=mass, g=acceleration of gravity), downward
- the buoyant force, equal to (
=mass of water displaced), upward
So the resultant force, which is the apparent weight of the rock (W'), is
which can be rewritten as
where m' is the apparent mass of the rock. Using:
m = 540 g
m' = 342 g
we find the mass of water displaced
(b)
If the rock is completely submerged, the volume of the rock corresponds to the volume of water displaced.
The volume of water displaced is given by
where
is the mass of the water displaced
is the density of the water
Substituting,
And so this is also the volume of the rock.
(c)
The average density of the rock is given by
where
m = 540 g = 0.540 kg is the mass of the rock
is its volume
Substituting into the equation, we find
Answer:
Approximately (assuming that the projectile was launched at angle of
above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing is the same as the altitude
at which this projectile was launched:
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is (upwards,) the vertical velocity right before landing would be
(downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be . In other words,
.
Hence, the time it takes to achieve a (vertical) velocity change of would be:
.
Hence, this projectile would be in the air for approximately .