Answer:
a) p = m1 v1 + m2 v2
, b) dp / dt = m1 a1 + m2 a2
, c) It is equivalent to force
dp / dt = 0
Explanation:
In this problem we have two blocks and the system is formed by the two bodies.
Part A. Initially they ask us to find the moment of the whole system
p = m1 v1 + m2 v2
Part B.
Find the derivative
dp / dt = m1 dv1dt + m2 dv2 / dt
dp / dt = m1 a1 + m2 a2
Part C.
Let's analyze the dimensions
m a = [kg] [m / s2] = [N]
It is equivalent to force
Part d
Acceleration is due to a net force applied
Part e
The acceleration of block 1 is due to the force exerted by block 2 during the moment change
Part f
Force of block 1 on block 2
True f12 = m1a1 f21 = m2a2
Part g
By the law of action and reaction are equal magnitude F12 = f21
Part H
dp / dt = 0
Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed
Because its a vacuum, there's no air resistance, they will fall at same time
Applying gravity acceleration rule g=9.8m/s which is taken as 10m/s sometimes.
Answer:
a) F = 1.26 10⁵ N, b) F = 2.255 10³ N, c) F_ {soil} = 3078 N
Explanation:
For this exercise we will use the relationship between momentum and moment
I = Δp
F t = p_f -p₀
a) with stiff legs, final speed is zero, initial velocity is down
Ft = 0-p₀
F = m v / t
let's calculate
F = 84.0 6.82 / 4.56 10⁻³
F = 1.26 10⁵ N
b) bending the legs
let's calculate
F = 84.0 6.82 / 0.254
F = 2.255 10³ N
c) It is requested to calculate the force of the ground on the man
∑ F = F_soil -W
F_soil = F + W
F_ {soil} = 2.255 103 + 84 9.8
F_ {soil} = 3078 N
Answer:
The ball travels a distance of 20 m in the time interval of 4 s
Explanation:
Using s = ut + 1/2at² where s = distance travelled by the ball, u = initial velocity of ball = 0 m/s (since it starts from rest), a = acceleration of the ball = 2.50 m/s² and t = time = 4 s.
Substituting the variables into the equation, we have
s = ut + 1/2at²
s = 0 × 4 s + 1/2 × 2.50 m/s² × (4 s)²
s = 0 + 1/2 × 2.50 m/s² × 16 s²
s = 1/2 × 40 m
s = 20 m
So, the ball travels a distance of 20 m in the time interval of 4 s.