Answer:
a) F = 1.26 10⁵ N, b) F = 2.44 10³ N, c) F_net = 1.82 10³ N directed vertically upwards
Explanation:
For this exercise we must use the relationship between momentum and momentum
I = Δp
F t = p_f -p₀
a) It asks to find the force
as the man stops the final velocity is zero
F = 0 - p₀ / t
the speed is directed downwards which is why it is negative, therefore the result is positive
F = m v₀ / t
F = 63.5 7.89 / 3.99 10⁻³
F = 1.26 10⁵ N
b) in this case flex the knees giving a time of t = 0.205 s
F = 63.5 7.89 / 0.205
F = 2.44 10³ N
c) The net force is
F_net = Sum F
F_net = F - W
F_net = F - mg
let's calculate
F_net = 2.44 10³ - 63.5 9.8
F_net = 1.82 10³ N
since it is positive it is directed vertically upwards
Answer:
Explanation:
the net force on the right left is 25 N and is directed upward
the net force on the left one is zero because 200N force act upward and 200N force act downward so both cancel each other and net force is zero
i hope this will help you
Answer: 17.03 rad/s^2
Explanation: One of the equation of motion that defines a circular motion with constant acceleration is given below as
ω = ω0 + αt
Where ω = final angular velocity = 9.3 rev/s
ω0 = initial angular velocity = 7.1 rev/s
α = angular acceleration = ?
t = time taken = 6.0s
By substituting the parameters, we have that
9.3 = 7.1 +α(6)
9.3 - 7.1 = 6α
16.3 = 6α
α = 16.3/6
α = 2.71 rev/s^2
We can also give the answer in rad/s^2 ( another unit of angular acceleration) by multipying by 2π
Hence, α = 2.71 × 2π , where π = 3.142
α = 17.03 rad/s^2