Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1
Answer:
v = 23.66 m/s
Explanation:
recall that one of the equations of motion may be expressed:
v² = u² + 2as,
Where
v = final velocity (we are asked to find this)
u = initial velocity = 0 m/s since we are told that it starts from rest
a = acceleration = 0.56m/s²
s = distance traveled = given as 500m
Simply substitute the known values into the equation:
v² = u² + 2as
v² = 0 + 2(0.56)(500)
v² = 560
v = √560
v = 23.66 m/s
Answer:
v = 5.166 10² m / s
Explanation:
We can solve this exercise using the kinematics equations
v = v₀ + at
as the bullet starts from rest its initial velocity is zero
v = a t
let's calculate
v = 6.3 10⁵ 8.2 10⁻⁴
v = 5.166 10² m / s
Answer:
Compaction and cementation
Explanation:
Cementation: As ions are deposited by fluids to form a compound that hardens loose sedimentary rocks.
Compaction: As the density of sedimentary rocks on edge of them are forced together through sediments.