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 .
Explanation:
Work = force × displacement
532 J = 48 N × d
d ≈ 11 m
Answer:
(B) 13.9 m
(C) 1.06 s
Explanation:
Given:
v₀ = 5.2 m/s
y₀ = 12.5 m
(A) The acceleration in free fall is -9.8 m/s².
(B) At maximum height, v = 0 m/s.
v² = v₀² + 2aΔy
(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)
y = 13.9 m
(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.
v = at + v₀
-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s
t = 1.06 s
Answer:
Due to application of heat on a saturated solution, the interparticle space increases due to increase in the kinetic energy of the particles which allows more solutes to dissolve in the solution thereby making it unsaturated.
